Essay Sample on Mathematics The System of Linear Equations

Essay Sample on Mathematics The System of Linear Equations PATTERNS WITHIN SYSTEM OF LINEAR EQUATION A system of linear equation is basically dealt with in the algebra unit. It is a collection of the linear equations involving variables of the same set in the in the equations that are involved. For example a 2Ãâ€"2 system of linear equations includes: x + 2y=10 3x + 4y=15 Here in both the cases the equations only involve two variables that is x and y and no other variable is included. In the example of a 33 system of linear equations it mostly includes the variables x, y and z for example; 2x + y-z =11 x- 2y + 2z =-2 3x-y+2z =5 Where only the three variables are involved There are also various properties of the patterns of the linear systems. We will start with the consistency property. If the systems of the equations have common solutions, then they are said to be consistent. This therefore means that graphically the lines should be straight lines. The independence property is also termed as the linear independence. The systems of equations are usually independent since to start with, they are derived algebraically from others. For example the system 3x+4y =9 and 6x +8y =18. There are different ways of solving the systems of linear equations that includes; The elimination of variables The substitution of variables technique The row reduction method The crammers’ rule The matrix method In the mathematical field, the general linear equation in the x and y is Ax+By=C where both the A and B in the equation are not zeros. The y-intercept in the line is the y-coordinate of that point where graphically, the non-vertical line that is drawn either manually or graphically intersects the y-axis. Also, the x-intercept is the point where the non horizontal line crosses the x-axis. Therefore the most general equation for a line with slope m and the y-intercept passing through b as the y intercept is written as y= mx + b. Therefore, one can easily find the slope and at the same time the y-intercept of any line. For example finding the slope and the y-intercept for 4x+5y=40 Solution: first and foremost, solve the equation for y to put it in the slope intercept format 4x+5y=20 5y=20-4x y=4-4/5x y=-4/5x+4 therefore the slope m=-4/5 and the y intercept is b=4 Consider this 2Ãâ€"2 system of linear equations 4x+3y=7 3x-2y=9 When we examine our first equation 4x+3y=7, there is a pattern in the constants of the equations used. Here 4 is the constant associated with the variable x and it therefore precedes the variable x. Also 3 is a constant that is preceding the variable y and the equation results to 7. In the second equation, 3x-2y=9, the constant 3 precedes our variable x and the constant -2 precedes the variable y making the equation to result to 9. It is also clear that in the two equations, the constants both have a difference of one. Solving the equations simultaneously, we first multiply the first equation by 3 and then multiply the second equation by 4 in order to eliminate the variable x and solve for the variable y. The equation then becomes; 12x+9y=21 12x-8y=36 17y=-15 Therefore solving the equation yields y=-15/17. Putting the value of x in any of the solution to obtain the value of x; 4(x) +3(-15/17) = 21 X=41/17 Graphically the system of equation is solved as This is first done by putting the two equations in the form of y=mx+b. The solution of the equations is by observing the point of intersection of the two lines that are plotted graphically. In this system of equation the solution therefore is (41/17, -15/17) Consider this 2Ãâ€"2 system of linear equations x+2y=3 and 2x-y=-4 The two equations are linear because the unknowns only appear to the first power, no unknown in the denominator of a fraction is in the equations and there are no products of unknowns. Therefore, the most general linear equation is a11x1 + a12x2 ++ a1nxn=b1 a21x1 +a22x2++a2nxn=b2 am1x1+ am2x1 +..+amnxn=bm m With unknowns x1, x2..xn and coefficients a1, a2an . In x+2y=3, the constant is 3 and the unknowns are x and y whereby x= 3-2y and y= (3-x) à ·2. The gradient of the linear equation is -1/2 and the y- intercept is 3/2. The gradient is negative therefore it is negatively sloped. In 2x-y =-4, the constant is -4 and the unknowns are x and y where x= (y-4) à ·2 and y=2x+4. The gradient of the line is 2 and the y- intercept is 4. The gradient is positive and therefore positively sloped. Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area [1 2: 3] and [2 -1: -4] The inverse is The solution therefore is; 1 2: x =3 2 -1: y =-4 1/5 2/5 3 =x 2/5 -1/5 -4 =y The graphic calculator here was used during this step to solve the matrix equation, normally if the equation is Ax= B then the solution is x= A-1B X=-1 y=2 The first function from the graph is sloped from left to right that is it is negatively sloped since the gradient is negative and the second equation is sloped from right to left since the gradient is positive. From the graph the solution of the equations is x and y2. This is read directly from the graph where the two lines intersect. In short, the solution to the system of equation is unique in that there is only one solution set to the system of equations and the solution satisfies the individual equations in the system of equations. Therefore, when x=-1 and y=2. Then -1+ (2Ãâ€"2) =3 and (2Ãâ€"-1) 2 = -4 which satisfy the equations that are given. Another example of linear equations is x+2y =3 3x-5y=9 This is a system of equations since it contains more than one equation. The solution set to the system of linear equation is the set of numbers n and m such that if we let x=n and y=m then we will obtain the result of the right hand side of the equation. For instance ax+by =c, if x=n and y=m then we obtain the result c given a and b are known constants. Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area 1 2: x =3 3 -5: y =9 The inverse of the equation is -5/11 -2/11 -3/11 1/11 The solution therefore is, -5/11 -2/11 3 =x -3/11 1/11 9 =y Therefore xand y The graph of the two equations is as shown below This is so since (3 1) + (2 0) =3 and (33)-(50) = 9 as proven from the equation. Now consider the 2 2 system below x2y=4 5x-y=1/5 In the first equation, x2y=4 f(x) = x/8. Therefore (0, 0) In the second equation, 5x-y=1/5 then f(x) = 5x-1/5. The gradient is 5 and the y intercept is -1/5 The matrix of the equation therefore is [1/8 -1: 0] [5 -1: 1/5] Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area -39/8 39/8: 0 = x -195/8 39/64: 1/5 = y Consider the following two by two system of equation y + 2x=7 y + x/2=3 In the first equation, the line is negatively sloped since y= f(x) = -2x+7. The gradient is -2 which means that the change in x compared to the change in y is -2. The y intercept is 7 and therefore when x=0, y=7 In the second equation, the equation line is also negatively sloped. The equation y= f(x) =-x/2 +3. The gradient is -1/2 and the y- intercept is 3 meaning when x=0, y=3 Each of the unknown variables can be solved using the graphical calculator in the matrix calculation area Therefore the solution to the equation is x=8/3 and y=5/3 The graph of f(x) =-2x+7 and f(x) =-x/2+3 are as follows: From the graph both the linear equations are negatively sloping but that of f(x)= -2x+1 is steeper than that of –x/2 + 3. The equations above are linear equations that results to linear curves and therefore two equations are enough to solve the equations. This two equations result into a square matrix. In the multiplication of matrices for instances, if A is an np matrix and B is a pm, then AB is the product of A and B denoted by AB and AB will be an nm matrix. That is AB exists if and only if number of columns of A is equal to the number of rows of B and that it should be noted that ABBA therefore the matrices do not commute. Therefore in solving the equation for example Ax=B, to find the values of x, the inverse of A is first found then multiplied by B. That is x= A-1B If A is a square matrix, we can find another matrix B known as the inverse of the matrix such that AB=BA=I. The inverse however can be a right inverse or a left inverse most commonly denoted as A-1. If AB=BA=I then A-1= B. Therefore if B exists then A is said to be invertible and non singular matrix. If B does not exist then A is said to be singular matrix. For the solution to be found in the equations, the matrix involved should be non singular. Theorem If A is a non singular matrix then A-1 is unique Proof Let A-1 =B then AB=BA=I Suppose B is not unique, then there exist C such that CB Then CA=AC=I But B=IB therefore (AC) B= (CA) B and thus C (AB) =CI =C Therefore B=C, a contradiction that B is unique An example of invertible matrix can be solved by looking at the following question. Solve the equations -4x-2y=8 and 6x+3y=12 The matrix of the equation is a 22 matrix and therefore the equation can written in the augmented form as shown Any matrix is said to be in reduced row echelon form if it satisfies the following conditions: Any row of all zeros appear at the bottom If a row does not consist of all zeros then its first non zero entity is called a leading 1 and it is one If any two successive rows the leading 1 of the lower is further to the right of the leading 1 of the highest row If a column contains a leading 1 then all the other entries are zero However, each of the unknown variables can be solved using the graphical calculator in the matrix calculation area But the inverse of the matrix does not exist since in fact the determinant is zero (-43) (6-2) =0. Therefore there is no solution to the equation above. Graphically, the lines to the equations are parallel and never intersect therefore there is no solution to the problems From the graph, it is clear that the two lines are parallel and are never to intersect and therefore this means that the equations do not have a solution. In some cases, the system will have many solutions in the algebraic sense, however geometrically, the lines will collide and look like there is only one line and therefore all the points along that line are indeed solutions to the equation. For example the equation -2x+y=8 and -4x+2y= 16 matrix to the equation is One of the solutions is x=0 and y=-4 and many other solutions. The matrix does not have an inverse as shown by the graphical calculator in the matrix calculation area since the determinant of the matrix is zero. This means that the solution of the equation is not one hence the equation has many solutions. In general, given any systems of linear equation with two unknown solution, the two lines will graphically intersect at one point. The point of intersection is the solution to the systems of linear equations. Also, the lines can be parallel to each other meaning that the system does not have any solution and finally the lines can collide and the solution to the system is not unique that is there are many solutions to the system of linear equations. The system with many solutions can be presented in the graph below using the equation given above. Note that: -4x+2y=16 2 {(-2x+y) =8} and therefore one equation is a multiple of the other which basically means that the equation is more or less the same. Remember that a system such as 2x-3y=7 and x+7y=11 can be written in the form = It is normally represented as Am=b where A= , b= and m= If b=0, then the system Am= b has m=0 as a trivial solution but if A-1 exists then m=0 is the only solution to the system. There is also the possibility of graphing those equations with piecewise defined functions. For instance there are functions such as |x| = In our example, we will graph one of the most commonly used piecewise defined functions. f (x) = In this case, the entire function is considered as one function in who’s the domain is the real numbers. APPLICATION OF LINEAR FUNCTIONS There are times when the solutions for the complicated functions cannot easily be obtained. This lead to the use of the linear equations that is the equations that are to only one degree to be used in the approximation of the complicated functions since they gives some little bit of accuracy and the linear functions are easy to work with. This is basically known as linearization. This is mostly used in conjunction with the differential functions. Here if the function, normally denoted as f is differentiable at x=a, then in this case the approximation function denoted as L(x) = f (a) +f’ (a) (x-a) is now what is known as the linearization of the function f at a. For example we will try to find the linearization of the function f(x) = at x=0 The above graph is now the linearization of the function at x=o and x=3. We now know that f’(x) =1/2(1+x)-1/2. We will therefore see that f (0) =1 and also that f’ (0) =1/2. Therefore this concludes that the linearization will therefore be: L(x) = 1+1/2(x-0) = 1+x/2 These are some of the general applications of the linear equations/functions and many other that are dealt with at the higher level of the course work Theorem in the solution of system of equation If A is invertible, then there is only one solution to Am=b which is the unique solution Proof Let w be any solution such that wA-1b That is Aw=b but since A is invertible A-1 exists that is Aw=b Therefore multiplying both sides to the left with A-1 we have A-1Aw = A-1b I w = A-1b w = A-1b which is a contradiction and therefore A-1b is the only solution to the system Next there are equations that are to the second degree and the linear equations are used to find the gradients at particular points through the use of the tangent line and the normal lines to the equations that are being considered in this case. These equations mostly include the parabolas and other quadratic equations among others. Though our main interest is not the parabolas and such equations, the linear equations are particularly used here to serve various mathematical purposes. The parabola for instance is a set that usually consist of all points in a plane that is equal in distance sense from a point that is given and also a given line. Mostly the parabolas will have a graph of equations of the form y= ax2+bx+c. We will for instance plot a graph of y=x2. In this case the graph is a simple graph that is curved in u shape. But mathematically, we may want to find the gradient of the graph at particular points. We will therefore use the current technology for graph plotting to plot both the graphs as shown; The tangent line is used to find the gradient of the curve at that particular point. The graph shown is a curve with the equation y=x2. The axis of the parabola is the y-axis that is it is the axis of symmetry.lso, the vertex of the parabola as seen from the graph is at the origin. The parabola is seen to open upwards when the values of the constant are positive and increasing and open up downwards if the values of the constant are negatively increasing. Now if we consider the 33 matrix system, there are 3 variables that are involved, we will concentrate on the variables x, y and z. For instance, let as consider the matrix below 2x+y-z=11: Here the constants are 2 that precede the variable x, 1 that precedes the variable y and -1 that precedes the variable z. The system can be solved using the usual matrix method, the elimination method or the use of a three dimension matrices. When we deal with the matrix method the graphing calculator here is used to find the inverse of the matrix. The system of equations can be basically being written as; M X=A Using the graphing calculator to find the inverse of the matrix will yield X= M-1A X= And therefore the solution to the equation becomes X = In this type of system, there are also the possibility of obtaining a unique solution, the; possibility of many solution and the option of no solution. The possibility of many solutions or no solution is as a result of having a singular matrix that is a matrix with a zero determinant. For example looking at the following system of solution x+2y+3z=4 4x+6y+8z=10 2x+y=-1 The determinant of the matrix is zero and therefore there can be the case where there are many solutions and graphically in a three dimension graph, the lines are common or the case where there are is no solution and the lines are parallel to each other. We can also use our technology to create a family of linear equations that are usually similar in characteristics. On the same set of the axis, we usually display the equations and evaluate them mathematically. The family of curves will include several lines which usually have a wide range of equations. This can be represented as; The family of linear equations above all have different gradients fro negative to zero to positive. In a 33 matrix, the solution can also be obtained geometrically and algebraically. This is so because the graph of the equations can be plotted in the graph especially with the current technologies and calculators and it can be done algebraically through various methods which include the elimination methods and the current modern methods. Therefore the 33 matrix can be dealt with in the same manner as the 22 matrix. There are many ways of proving mathematical theorems and terms such as the contradiction method, proving by induction and many others in the above matrix we have used the contradiction method. In the 33 matrix, we are going to basically see how to prove by induction the conjectures that are involved. Conjectures in mathematics are some of the propositions and they are easily not disapproved since they are believed to be true For instance the sequence an= n (n-1) is the sequence such that a1= 1*0 a2= 2*1 : : an= n*(n-1) When we sum up the sequence of the first n numbers we obtain a series and therefore sn= a1+a2++an. Therefore; S1=0 S2=2 Sn= Sn-1+an Next, the difference between successive sums is made until the constant term in the series is obtained so long as the nth term n0, This will result to a polynomial of the third degree in order for the constant terms to be obtained in that the equation for the series will therefore be Sn = Ax3+Bx2+Cx+D where A,B,C and D are constants that are and xâ ± ¤ Now replacing x in the equation with the natural numbers 1,2,3,4,5.. we get the A+B+C+D=0 8A+4B+2C+D=2 27A+9B+3C+D=8 64A+16B+4C+D=20 This is a four equation system since there are four different unknown variables. Therefore we will use the graphing calculator to find the solutions to the unknown variables 1 1 1 1: 0 8 4 2 1: 2 27 9 3 1: 8 64 16 4 1: 20 We will find that A=1/3, B=0, C=-1/3 and D=0 Sn=1/3 x3-1/3 x The graph for the equation is therefore as follows Proving the equation by the induction method therefore will be (for n0); For n=1: =1/3*13 1/3* 1 =0 For n=3: =1/3*33 -1/3*3 =8 For n=5: =1/3*53 1/3*5 =40 Therefore we can assume that the equation is true for all values of natural number that is n0, We therefore assume that the equation is true for n=k Therefore for n=n+1, Sk+1= Sk+ (k+1)*k =1/3k3 -1/3k +k2 +k =1/3(k+1)3 -1/3(k+1) Since the expression is true for n=k+1 is true, the equation is true by induction. In the mathematical sense, a function of a polynomial p is normally written as p(x) =anxn+an-1+.+a1 x+ a0. In this case the n are non negative integers and the a’s are the coefficients of the polynomial itself. Usually all the polynomials have the domain of (-, ). In this case we can say that the linear functions themselves are polynomials of degree one while the quadratic functions are polynomials of the second degree and so on. As with our 33 matrix, the polynomial involved was a cubic function of the third degree. For instance the polynomial y=84-143-92+11x-1 Linear Algebraic Equations A teacher is looking for the best option in purchasing school supplies for a classroom. Company A is offering a discount for every dollar amount spent; Company B is offering a higher discount for every dollar spent above $20. Determine which company will offer a better price based upon the dollar amount the teacher spends on the school. In this scenario, it mostly involves the computation of the purchase of the school inventories at a cheaper price. Inventory generally is the stock of raw materials, work in progress units, finished goods, consumables and spare parts being held in store at a given time period. There are different kinds and groups of inventories that includes; movement inventories which are inventories on transit from one point to another, safety stock or the buffer stock which are the inventories that must always be maintained in the store so as to meet the unexpected demand, cyclical inventory, anticipatory inventory and the decoupling inventory. However, in this scenario we will focus on how to purchase inventory while at the same time using the mathematical knowledge to reduce inventory related costs. Here, the teacher is looking for the best option in purchasing school supplies for a classroom and therefore the best option is the option with reduced costs. The customer also has to ensure that tho ugh the goods are purchased at a cheaper price, they are of the best and desirable quality. The supplier of the goods should also be in a position to supply goods to the customer when they are needed both in the short term period and in the long term period and in time as to the date of the specifications. Therefore the customer has to look deep into these needs before making the decision on where to make their orders. The hypothetical customer, the teacher in this case has to make an informed decision based on questions such as how many units to order at that time, how often should the school supplies be made, how many orders are to be placed in that particular year and this is mainly done to reduce cost. In this scenario therefore we will focus mainly on two cost options that are for Company A which is offering a discount for every dollar spent. This is where a constant rate of discount for every dollar spent. This is where a constant rate of discount is offered irrespective of the number of units purchased and it is commonly known as a single discount. We will assume that the unit price of each product that is to be purchased is $5 and that the discount for every dollar spent is 5 percent (5%). In this scenario, the discount offered is for every dollar that will be spent and no conditions as to the amount and the limit of expenditure. The second option is for Company B which is offering a higher discount for every dollar spent above $20. This therefore guarantees the teacher discount after spending $20 in the purchase of school supplies which will be a much higher discount than that of the purchase of the goods worth $20. In this case the teacher will get a discount similar to that of the single discount up to the expenditure of $20 and later the discount is increased accordingly. We will therefore assume that the unit purchase price is $5 and that every dollar spent to the expenditure of $20 is 5%, with more dollars spent, the discount increases to 7.5%. In our scenario, the demand should be known in advance with certainty and will remain constant within the relevant range. The algebraic equations to represent the cost of each option are: Company A: Offers a discount for every dollar spent Here there are many cost related to the purchase of the school supplies which includes the purchase cost, the ordering cost, the holding cost and in some cases the shortage cost. However we will only focus on the purchase cost and ignore all the other related cost inorder to come up with the required linear equations. Let’s assume further that the teacher purchases x units of the school supplies Unit price = $5 Discount =5% Let the total cost=y Total purchase cost =$5 * (100% -5%) *x Total purchase cost = $5 *0.95 *x = 4.75x Company B: Offers a higher discount of 7.55 for every dollar spent above $20. The teacher here should know that for the first $20 spent, the discount is 5% and above the expenditure of $20, the discount increases to 7.5%. This will probably lure customers desire to purchase more but we will try evaluating the two equations. The equation for company B is therefore as follows Let’s assume that the teacher purchases x units of the school supplies The discount for the first 4 purchases of the school supplies =5% i.e ($20/5) Unit price =$5 Let the total cost=y The discount for the purchase of more than 4 =7.5% Total purchase cost = ($5*4*0.95) + ($5*0.925)(x-4) Total purchase cost = $19+4.625x –$18.5 =$0.5+4.625x The equation of company A is used since the company only offers a single discount for all the purchases that are made by the customer. Therefore the discount will be distributed equally. For Company B, there is a constant in the algebraic equation since in the purchase of the first $20 items, the discount is 5% that is it is constant and since the customer has to purchase more than this to gain the discount of 7.5% then that part of the equation will vary with the extra units purchased. The solution to the equation can be done through several ways such as elimination method, substitution method or the graphical method; the equations are normally written as y= 4.75x y= 0.5+4.625x We are going to solve the equation using the substitution method. Since in the first equation y=4.75x, we will substitute this to the second equation.4.75x= 0.5+4.625x and we therefore solve the equation mathematically. In this equation, the solution is x=4. Where the total costs will be the same. But with the increase in the purchase of the school supplies, the total cost will be higher for the purchase related to Company A than that of the purchase from Company B. Also, with the decrease in the purchase of the school supplies, the cost purchases from Company A are less than that of Company B. The graph of the two scenarios can be represented as follows. Though the cost associated in the two scenarios are close, there is a negligible difference as a result of the discounts. It is therefore correct to conclude that if the teacher is in need of less than four units of purchase, it is advisable to purchase the school supplies from Company A, if the teacher wants to purchase 4 units of item, this can be done from any company and if it is more than 4 company, it is cost effective to purchase from Company B. At the lower levels of purchase, presence of discount appears attractive for the company with a single discount for any unit of purchase made. That is, there is no constant related to the purchase of goods in the algebraic equation.. However beyond a certain level of purchase, taking up a discount results into a net increase in total cost in the company using a single discount method. In the other case for company with an increased discount after purchase of some discount, the purchase of many items become cost effective in this company. Therefore the teacher should ensure that he or she takes up the least quantity required to qualify for the highest discount in order for the total cost to be less than that of company A. Since the principle of discount states that only the least quantity required to qualify for the discount should be purchased.

Algebraic Operations on ACT Math Strategies and Formulas

Algebraic Operations on ACT Math Strategies and Formulas SAT / ACT Prep Online Guides and Tips Variables, exponents, and more variables, whoo! ACT operations questions will involve all of these (and so much more!). So if you ever wondered what to do with or how to solve some of those extra long and clunky algebra problems (â€Å"What is the equivalent to ${2/3}a^2b - (18b - 6c) +$ †¦Ã¢â‚¬  you get the picture), then this is the guide for you. This will be your complete guide to ACT operations questions- what they’ll look like on the test, how to perform operations with multiple variables and exponents, and what kinds of methods and strategies you’ll need to get them done as fast and as accurately as possible. You'll see these types of questions at least three times on any given ACT, so let's take a look. What Are Operations? There are four basic mathematical operations- adding, subtracting, multiplying, and dividing. The end goal for any particular algebra problem may be different, depending on the question, but the operations and the methods to solve them will be the same. For example, when solving a single variable equation or a system of equations, your ultimate objective is to solve for a missing variable. However, when solving an ACT operations problem, you must use your knowledge of mathematical operations to identify an equivalent expression (NOT solve for a missing variable). This means that the answer to these types of problems will always include a variable or multiple variables, since we are not actually finding the value of the variable. Let’s look at two examples, side-by-side. This is a single variable equation. Your objective is to find $x$. If $(9x-9)=-$, then $x=$? A. $-{92/9}$B. $-{20/9}$C. $-{/9}$D. $-{2/9}$E. $70/9$ This is an ACT operations problem. You must find an equivalent expression after performing a mathematical operation on a polynomial. The product $(2x^4y)(3x^5y^8)$ is equivalent to: F. $5x^9y^9$G. $6x^9y^8$H. $6x^9y^9$J. $5x^{20}y^8$K. $6x^{20}y^8$ (We will go through exactly how to solve this problem shortly) Let's break down each component of an operations problem, step-by-step. (Also, bonus French braid lesson!) Operation Question How-To's Let us look at how to identify operations questions when you see them and how to solve for your answer. How to Identify an Operations Problem As we said before, the end goal of an operations problem is not to solve for a missing variable. Because of this, you can identify an operations problem by looking at your answer choices. If the question involves variables (instead of integers) in the given equation and in the answer choices, then it is likely you are dealing with an operations problem. This means that if the problem asks you to identify an â€Å"equivalent† expression or the â€Å"simplified form† of an expression, then it is highly likely that you are dealing with an operations problem. How to Solve an Operations Problem In order to solve these types of questions, you have two options: you can either solve your problems by using algebra, or by using the strategy of plugging in numbers. Let’s begin by looking at how algebraic operations work. First, you must understand how to add, multiply, subtract, and divide terms with variables and exponents. (Before we go through how to do this, be sure to brush up on your understanding of exponents and integers.) So let us look at the rules of how to manipulate terms with variables and exponents. Addition and Subtraction When adding or subtracting terms with variables (and/or exponents), you can only add or subtract terms that have the exact same variable. This rule includes variables with exponents- only terms with variables raised to the same power may be added together (or subtracted). For example, $x$ and $x^2$ CANNOT be combined into one term (i.e. $2x^2$ or $x^3$). It can only be written as $x + x^2$. To add terms with variables and/or exponents, simply add the numbers before the variable (the coefficients) just as you would add any numbers without variables, and keep the variables intact. (Note: if there is no coefficient in front of the variable, it is worth 1. $x$ is the same thing as $1x$.) Again, if one term has an additional variable or is raised to a different power, the two terms cannot be added together. Yes: $x + 4x = 5x$ $10xy - 2xy = 8xy$ No: $6x + 5y$ $xy - 2x - y$ $x + x^2 + x^3$ These expressions all have terms with different variables (or variables to different powers) and so CANNOT be combined into one term. How they are written above is as simplified as they can ever get. Multiplication and Division When multiplying terms with variables, you may multiply any variable term with another. The variables do not have to match in order for you to multiply the terms- the variables instead are combined, or taken to an additional exponent if the variables are the same, after multiplying. (For more on multiplying numbers with exponents, check out the section on exponents in our guide to advanced integers) $x * y = xy$ $ab * c = abc$ $z * z = z^2$ The variables in front of the terms (the coefficients) are also multiplied with one another as usual. This new coefficient will then be attached to the combined variables. $2x * 3y = 6xy$ $3ab * c = 3abc$ Just as when we multiplying variable terms, we must take each component separately when we divide them. This means that the coefficients will be reduced/divided with regard to one another (just as with regular division), as will the variables. (Note: again, if your variables involve exponents, now might be a good time to brush up on your rules of dividing with exponents.) $${8xy}/{2x} = 4y$$ $${5a^2b^3}/{15a^2b^2} = b/3$$ $${30y + 45}/5 = 6y + 9$$ When working on operations problems, first take each component separately, before you put them together. Typical Operation Questions Though there are several ways an operations question may be presented to you on the ACT, the principles behind each problem are essentially the same- you must manipulate terms with variables by performing one (or more) of the four mathematical operations on them. Most of the operations problems you’ll see on the ACT will ask you to perform a mathematical operation (subtraction, addition, multiplication, or division) on a term or expression with variables and then ask you to identify the â€Å"equivalent† expression in the answer choices. More rarely, the question may ask you to manipulate an expression in order to present your equation â€Å"in terms of† another variable (e.g. â€Å"which of the following expressions shows the equation in terms of $x$?†). Now let’s look at the different kinds of operations problems in action. The product $(2x^4y)(3x^5y^8)$ is equivalent to: F. $5x^9y^9$G. $6x^9y^8$H. $6x^9y^9$J. $5x^{20}y^8$K. $6x^{20}y^8$ Here, we have our problem from earlier, but now we know how to go about solving it using algebra. We also have a second method for solving the question (for those of you are uninterested in or unwilling to use algebra), and that is to use the strategy of plugging in numbers. We’ll look at each method in turn. Solving Method 1: Algebra operations Knowing what we know about algebraic operations, we can multiply our terms. First, we must multiply our coefficients: $2 * 3 = 6$ This will be the coefficient in front of our new term, so we can eliminate answer choices F and J. Next, let us multiply our individual variables. $x^4 * x^5$ $x^[4 + 5]$ $x^9$ And, finally, our last variable. $y * y^8$ $y^[1 + 8]$ $y^9$ Now, combine each piece of our term to find our final answer: $6{x^9}y^9$ Our final answer is H, $6{x^9}y^9$ Solving Method 2: Plugging in our own numbers Alternatively, we can find our answer by plugging in our own numbers (remember- any time the question uses variables, we can plug in our own numbers). Let us say that $x = 2$ and $y = 3$ (Why those numbers? Why not! Any numbers will do- except for 1 or 0, which is explained in our PIN guide- but since we are working with exponents, smaller numbers will give us more manageable results.) So let us look at our first term and convert it into an integer using the numbers we selected to replace our variables. $2{x^4}y$ $2(2^4)(3)$ $2(16)(3)$ $96$ Now, let us do the same to our second term. $3{x^5}{y^8}$ $3(2^5)(3^8)$ $3(32)(6,561)$ $629,856$ And finally, we must multiply our terms together. $(2{x^4}y)(3{x^5}{y^8})$ $(96)(629,856)$ $60,466,176$ Now, we need to find the answer in our answer choices that matches our result. We must plug in our same values for $x$ and $y$ as we did here and then see which answer choice gives us the same result. If you are familiar with the process of using PIN, you know that our best option is usually to start with the middle answer choice. So let us test answer choice H to start. $6{x^9}y^9$ $6(2^9)(3^9)$ $6(512)(19,683)$ $60,466,176$ Success! We have found our correct answer on the first try! (Note: if our first option had not worked, we would have seen whether it was too low or too high and then picked our next answer choice to try, accordingly.) Our final answer is again H, $6{x^9}y^9$ Now let us look at our second type of problem. For all real numbers $b$ and $c$ such that the product of $c$ and 3 is $b$, which of the following expressions represents the sum of $c$ and 3 in terms of $b$? A. $b+3$B. $3b+3$C. $3(b+3)$D. ${b+3}/3$E. $b/3+3$ This question requires us to translate the problem first into an equation. Then, we must manipulate that equation until we have isolated a different variable than the original. Again, we have two methods with which to solve this question: algebra or PIN. Let us look at both. Solving Method 1: Algebra First, let us begin by translating our equation into an algebraic one. We are told that the product of $c$ and 3 is equal to $b$. A â€Å"product† means we must multiply $c$ and 3 and so our equation looks like this: $3c = b$ Now we are asked to find the sum of $c$ and 3. This means we must isolate $c$ so that we can add them together. So let us first isolate $c$ by using our knowledge of algebraic operations. $3c = b$ $c = b/3$ Now, we can sum $c$ and 3 by replacing our $c$ with $b/3$. $c + 3$ ${b/3} + 3$ Our answer matches answer choice E. Our final answer is E. Solving Method 2: Plugging in numbers Alternatively, we can use our technique of plugging in numbers. Because our question deals with variables, we can choose our own numbers (so long as they follow the rules of our given information.) We are told that the product of $c$ and 3 is equal to $b$. So let us assign a value to $c$ and use this information to find the value of $b$. So let us say that $c = 4$. (Why 4? Why not!) If $c = 4$, then the product of $c$ and 3 is: $3c = b$ $3(4) = b$ $b = 12$ So, when $c$ equals 4, $b$ equals 12. Now we must find the sum of $c$ and 3. $3 + c$ $3 + (4)$ $7$ Now that we have found our sum, we must identify the answer choice that gives us this sum. All of our answer choices are presented to us in terms of $b$, so we will use our found value of 12 to replace $b$ for each. As with all PIN questions, let us start with the middle answer option. Answer choice C gives us: $3(b + 3)$ We can tell just by looking at it that this will be far larger than 7, but we can always test this out. $3(12 + 3)$ $3(15)$ $45$ We can eliminate answer choice C. Just by glancing, we can see that answer choices A and B will also be larger than 12, which means we can eliminate them as well. Let us try answer choice D. ${b + 3}/3$ ${12 + 3}/3$ $15/3$ $5$ Answer choice D did not match our sum, which means we can eliminate it as well. By process of elimination, we are left with answer choice E, but let us test it to be sure. ${b/3} + 3$ ${12/3} + 3$ $4 + 3$ $7$ Success! We have found the answer choice that matches the sum we found. Our final answer is, once again, E, ${b/3} + 3$. As you can see, the answer to your operations questions will always be in variables and the problem will always require you to interpret and manipulate expressions with variables, but there are always multiple options for how to solve these types of problems. You've got the power to decide how you would like to solve and manipulate your operations problems. Magic! Strategies for Solving Operations Questions Now that we’ve seen the types of operations questions you’ll see on the ACT, let’s review our solving strategies. #1: Use PIN when needed (or to double-check your answer) If you ever feel concerned that you may be going down the wrong path while manipulating your operations problems, or if you simply want to double-check your answer, it's never a bad idea to use the strategy of plugging in numbers. Although it can take a little longer plug in your own numbers for your variables, you'll never have to fear misremembering how to manipulate your exponents, your variables, or your equations as a whole. Once you're able to use real numbers for your variables, the math will be a piece of cake. #2: Focus on one aspect of the term at a time It can become all too easy to lose yourself when working with multiple variables at once, especially when it comes to multiplication and division. The test-makers know this and will provide bait answers for any number of common mistakes. In order to keep all your components organized, focus on just one piece of each expression at a time. First, look at the coefficients, then look at the variables. This will help keep all your moving pieces in order and lessen the odds of mix-ups and mistakes. #3: Eliminate your answer options as you go Operations problems can sometimes mess with your head, not because they are inherently difficult, but because the ACT is a marathon and your brain can get tired and confused (and lazy). This, combined with the fact that all the answer choices generally look quite similar, with only small differences- a minus sign instead of a plus sign, one coefficient difference, etc.- can lead you to select the wrong answer, even when you know what the correct one should be. To avoid this kind of careless error (the worst kind of error!), eliminate your answer choices as you go through your problem. Know that the coefficient for your $y$ value must be 3? Immediately cross out any answer choices that give you anything other than $3y$. It may seem inefficient to solve problems this way, but it will keep your answers much more clear. #4: Keep careful track of your negatives Not only can it be difficult to keep track of multiple variables, but it's even easier to mix-up the proper negative and positive signs. Many students make careless errors with their negative signs and the ACT test-makers are all too aware of this. They will provide all manner of bait answers for anyone who misplaces even a single negative sign, so be very careful. $(a+2b+3c)-(4a+6b-5c)$ is equivalent to: A. $-4a-8b-2c$B. $-4a-4b+8c$C. $-3a+8b-2c$D. $-3a-4b-2c$E. $-3a-4b+8c$ For a problem like this, we are being asked to subtract the entire expression, $4a + 6b - 5c$, from the entire expression, $a + 2b + 3c$. This means that the negative sign will be negating every term in the expression $4a + 6b - 5c$. So we must put a negative sign in front of each term. $4a$ becomes $-4a$ $6b$ becomes $-6b$ $-5c$ becomes $- -5c$ or $+5c$. Now let us put these pieces together with the first expression. $a - 4a = -3a$ $2b - 6b = -4b$ $3c + 5c = 8c$ Our final expression will be: $-3a - 4b + 8c$ Our final answer is E, $-3a - 4b + 8c$. [Note: many (many!) students put a negative sign only in front of the first term in the parenthesis, which in this case the $4a$. If you had done this, you would have gotten: $a - 4a = -3a$ $2b + 6b = 8b$ $3c - 5c = -2c$. This would have given you answer choice C, $-3a + 8b - 2c$. Again the test-makers know this is a common error and there will always be a bait answer to tempt anyone who makes this kind of mistake.] Operations in the "real world." Hyuk, yuk, yuk. Test Your Knowledge Now that we’ve gone through the tips and tricks of operations questions, it’s time to put your knowledge to the test with more real ACT math problems. 1. Which of the following is an equivalent simplified expression for $2(4x+7)-3(2x-4)$? F. $x+2$G. $2x + 2$H. $2x+26$J. $3x+10$K. $3x+$ 2.Which of the following expressions is equivalent to ${1/2}y^2(6x+2y+12x-2y)$? A. $9xy^2$B. $18xy$C. $3xy^2 + 12x$D. $9xy^2-2y^3$E. $3xy^2+12x-y^3-2y$ 3.$t^2-59t+54-82t^2+60t$ is equivalent to: F. $-26t^2$G. $-26t^6$H. $-81t^4+t^2+54$J. $-81t^2+t+54$K. $-82t^2+t+54$ 4.The expression $-8x^3(7x^6-3x^5)$ is equivalent to: A. $-56x^9+24x^8$B. $-56x^9-24x^8$C. $-56x^18+24x^15$D. $-56^18-24x^15$E. $-32x^4$ Answers: H, A, J, A Answer Explanations: 1. As always, we can solve this question using algebra or using PIN. Let us look at both ways. Method 1: Algebra First, we must distribute out our terms. Only afterwards will we subtract them. Let us take each half of our expression by itself. $2(4x + 7)$ $8x + 14$ $ -3(2x - 4)$ $-6x + 12$ (Note: keep careful track of your negatives here, especially in the second half of our expression.) Now, we can put the two together. $8x + 14 - 6x + 12$ $2x + 26$ We cannot go any further, as we have combined all our like terms. Our final answer is H, $2x + 26$ Method 2: PIN As an alternative to algebra, we can always use plugging in numbers. So let us assign our own value to $x$, which we will call 3. (Why 3? Why not!) This means that we will replace any $x$ in our given equation with a 3. $2(4x + 7) - 3(2x - 4)$ $2(4(3) + 7) - 3(2(3) - 4)$ $2(12 + 7) - 3(6 - 4)$ $2(19) - 3(2)$ $38 - 6$ $32$ Now, let us find the answer choice that matches with our found answer of 32, once we replace the $x$ with 3. As usual, when using PIN, let us start with the middle answer option. $2x + 26$ $2(3) +26$ $6 + 26$ $32$ Success! We found our answer on the first try. But remember- when using PIN, always check your other answer options to make sure there are not repeat correct answers. We can see straightaway that answer choices F and G will be too small, since answer choice H was a match. So let us try answer choice J. $3x + 10$ $3(3) + 10$ $9 + 10$ $19$ This answer choice is too small and we can see just by looking that answer choice K will be too small as well (since they only differ by 1). This means we are safe with our answer choice H, as no others produced a match. Our final answer is H, $2x + 26$. As we saw from earlier in the guide and from the example problem above, we can always use algebra or PIN for our operations problems. Knowing that, we will only go through one method each for the rest of our answer explanations. 2: For this problem, let us do our solve using algebra (again, we could also use PIN, but for the sake of brevity, we are only choosing one method for each problem). We are given the equation: ${1/2}y^2(6x + 2y + 12x - 2y)$ Now, let us first make life easier by combining the like terms in the parenthesis. $(6x + 2y + 12x - 2y)$ $(6x + 12x + 2y - 2y)$ $(18x)$ The $y$ terms cancel one another out, so we are left with only $18x$ in the parenthesis. Now, we must multiply our $18x$ by ${1/2}y^2$. As always, when multiplying, we must multiply first the coefficients and then combine them with the combined variables. So: ${1/2}y^2 * 18x$ $(1/2) * 18 = 9$ $y^2 * x = y^2x$ Put the two together and we have: $9y^2x$ So our final answer is A, $9xy^2$ 3: Because we used algebra last time, let us try our hand at solving this question using PIN. Because we are using our own numbers, we don’t have to worry about whether or not we are matching up the right terms, or if we are combining them incorrectly; we can bypass all the mess and use numbers instead. We have one variable, $t$, so let us say that $t = 2$. (Why 2? As always, why not!) $t^2 - 59t + 54 - 82t^2 + 60t$ $(2)^2 - 59(2) + 54 - 82(2)^2 + 60(t)$ $4 - 8 + 54 - 328 + 120$ $-268$ Now, we must find the answer choice that matches our found answer of 102, once we replace $t$ with 2. Let us start in the middle, with answer choice H. $-81t^4 + t^2 + 54$ $-81(2)^4 + (2)^2 + 54$ $-81(16) + 4 + 54$ $-1296 + 58$ $-1238$ We can see just by looking that answer choice G will be too small as well ($-26 * 16 = -416$), and answer choice F will be too large (-26 * 4 = -104). So let us try answer choice J. $-81t^2 + t + 54$ $-81(2)^2 + 2 + 54$ $-81(4) + 56$ $-324 + 56$ $-268$ Success! And we can also see that the only difference between answer choices J and K are the coefficient in front of $t^2$ (-81 vs. -82), so we know that answer K would produce an incorrect and smaller number than answer choice J. Our final answer is J, $-81t^2 + t + 54$ 4: Because we used PIN last time, let us use algebra for this problem. Because we do not have like terms in the parenthesis, we must distribute out our expression using multiplication. $-8x^3(7x^6 - 3x^5)$ $-8x^3(7x^6) - -8x^3(3x^5)$ And take each piece separately. $-8x^3(7x^6)$ = $-8 * 7 = -56$ and $x^3 * x^6 = x^[3 + 6] = x^9$ (for more on this, look to the section on exponents in our advanced integers guide). So, combined, we have: $-56x^9$ And the other half of our expression will be the same. $- -8x^3(3x^5)$ $8x^3(3x^5)$ = $8 * 3 = 24$ and $x^3 * x^5 = x^[3 + 5] = x^8$ So, combined, we have: $24x^8$ Now our equation looks like this: $-56x^9 + 24x^8$ Our final answer is A, $-56x^9 + 24x^8$ (Take care! The only difference between answer choice A and B is the negative sign. If you weren’t careful with your double negatives, you may have fallen for this bait answer.) Ten thousand gold stars for solving your operations problems! The Take-Aways Though operations problems are easy to get wrong if you’re going too quickly through the test (or trying to solve them in your head), the basic elements are the same as any problem with variables- combine like terms, keep your work organized, and use PIN if you feel overwhelmed (or simply want to double-check your answer). You have a multitude of options for solving ACT algebra questions, so don’t be afraid to use them. What’s Next? Still in the mood for math? Well we've got you covered! First, take a gander at exactly what's tested on the ACT math section in order to get a feel for your strong and weak points. Next, dive right into our ACT math guides for any topic you feel you haven't quite mastered (or just any topic you want to refresh). From circles to ratios, slopes to polygons, we've got your back. Running out of time on the ACT math section? Check out our guide on how to help maximize your avaialable time in order to get your best score possible. Nervous about test day? Ease your mind by taking a look at what to do the night before and the day of the test. Trying for a perfect score? Look no further than our guide to getting a perfect 36 on the ACT math, written by a perfect-scorer. Want to improve your ACT score by 4 points? Check out our best-in-class online ACT prep program. We guarantee your money back if you don't improve your ACT score by 4 points or more. Our program is entirely online, and it customizes what you study to your strengths and weaknesses. If you liked this Math lesson, you'll love our program. Along with more detailed lessons, you'll get thousands of practice problems organized by individual skills so you learn most effectively. We'll also give you a step-by-step program to follow so you'll never be confused about what to study next. Check out our 5-day free trial: