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Exercises 1 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the graph, we can tell that this point is (3,1).Let's check this solution by substituting the point into each equation. Note that in order to verify the answer not only the point has to satisfy one of the equations but also the other one. {y=-31​x+2y=x−2​(I)(II)​(I), (II): x=3, y=1{1=?-31​⋅3+21=?3−2​(I):  3a​⋅3=a{1=?-1+21=?3−2​(I), (II):  Add and subtract terms{1=11=1​ Since both equations hold true after substituting the point, (3,1) is a solution of the system of equations.
Exercises 2 The solution of the system of the linear equations is the point where the graphs of the equations intersect. From the given graph, it seems like this point is (-2,-2).We can check this solution by substituting the point into each equation.First equation Let's substitute (-2,-2) into the first equation and simplify. y=21​x−1x=-2, y=-2-2=?21​(-2)−1a(-b)=-a⋅b-2=?-21​⋅2−12a​⋅2=a-2=?-1−1Subtract term-2=-2 The equation is true.Second equation Let's now do the same with the second equation. y=4x+6x=-2, y=-2-2=?4⋅(-2)+6a(-b)=-a⋅b-2=?-8+6Add terms-2=-2 The second equation is also true.Conclusion Since both equations hold true after substituting the point, it is the solution of the system of equations.
Exercises 3 The solution of the system of the linear equations is the point where the graphs intersect. From the graph, we can tell that this point is (0,1).Let's check this solution by substituting the point into each equation. Note that in order to verify the answer, the point has to satisfy both of the equations. {y=1y=2x+1​(I)(II)​(I), (II): x=0, y=1{1=11=?2(0)+1​(II):  Zero Property of Multiplication{1=11=1​ Since both equations hold true after substituting the point, we can confirm that (0,1) is a solution for the system of equations.
Exercises 4 When solving a system of equations using substitution, there are three steps.Isolate a variable in one of the equations. Substitute the expression for that variable into the other equation and solve. Substitute this solution into one of the equations and solve for the value of the other variable. For this exercise, y is already isolated in one equation, so we can skip straight to step 2! {y=x−4-2x+y=18​(I)(II)​(II): y=x−4{y=x−4-2x+x−4=18​(II): Add terms{y=x−4-x−4=18​(II): LHS+4=RHS+4{y=x−4-x=22​(II):  Change signs{y=x−4x=-22​ Now, to find the value of y, we need to substitute x=-22 into either one of the equations in the given system. Let's use the first equation. {y=x−4x=-22​(I): x=-22{y=-22−4x=-22​(I):  Subtract term{y=-26x=-22​ The solution, or point of intersection, to this system of equations is the point (-22,-26).Checking Our Answer To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct. {y=x−4-2x+y=18​(I)(II)​(I), (II):  x=-22, y=-26{-26=?-22−4-2(-22)+(-26)=?18​(II):  -a(-b)=a⋅b{-26=?-22−444−26=?18​(I), (II):  Subtract terms{-26=-2618=18​ Because both equations are true statements, we know that our solution is correct.
Exercises 5 In order to use the Substitution Method to solve a system of equations we need to begin by isolating either variable in one of the equations. We can then substitute the expression for that variable into the second equation. We have been given the following system of equations. {2y+x=-4y−x=-5​(I)(II)​ We will start by isolating the y variable in Equation (II). y−x=-5⇔y=-5+x Now we can substitute (II) into (I) and solve for x. {2y+x=-4y=-5+x​(I)(II)​(I): y=-5+x{2(-5+x)+x=-4y=-5+x​(I): Distribute 2{-10+2x+x=-4y=-5+x​(I): Add terms{-10+3x=-4y=-5+x​(I): LHS+10=RHS+10{3x=6y=-5+x​(I): LHS/3=RHS/3{x=2y=-5+x​ We can now substitute x=2 into Equation (II) to solve for y. {x=2y=-5+x​(II): x=2{x=2y=-5+2​(II):  Add terms{x=2y=-3​ The solution to the system is (2,-3).Checking the solution In order to check the solution we need to substitute x by 2 and y by -3 in both equations and verify we end up in an identity. {2y+x=-4y−x=-5​(I)(II)​(I), (II): x=2, y=-3{2(-3)+2=?-4-3−2=?-5​(I): a(-b)=-a⋅b{-6+2=?-4-3−2=?-5​(I), (II): Add and subtract terms{-4=-4-5=-5​ The solution verifies both equations.
Exercises 6 Using the Substitution Method to solve a system of equations requires us to begin by isolating either variable in one of the equations. We can then substitute the expression for that variable into the second equation. We have been given the following system of equations. {3x−5y=13x+4y=10​(I)(II)​ Before we can substitute one equation into the other, we must alter one of the equations so that it is formatted as x=… or y=…Finding y Since the coefficient of x in (II) is 1, it will only require one step to isolate x. x+4y=10⇔x=10−4y Now we can substitute (II) into (I) and solve for y. {3x−5y=13x=10−4y​(I)(II)​(I): x=10−4y{3(10−4y)−5y=13x=10−4y​(I): Distribute 3{30−12y−5y=13x=10−4y​(I): Subtract term{30−17y=13x=10−4y​(I): LHS−30=RHS−30{-17y=-17x=10−4y​(I):  LHS/(-17)=RHS/(-17){y=1x=10−4y​ The y part of our solution is y=1.Finding x We can now substitute y=1 into either (I) or (II) to solve for x. Let's use (II). {y=1x=10−4y​(I)(II)​(II): y=1{y=1x=10−4⋅1​(II):  a⋅1=a{y=1x=10−4​(II):  Subtract term{y=1x=6​ The x part of our solution is x=6, so the solution to the system is (6,1).Checking Our Answer To check our answer, we will substitute the solution we found into both equations. If both equations remain true, our solution is correct. {3x−5y=13x=10−4y​(I)(II)​x=6, y=1{3(6)−5(1)=13(6)=10−4(1)​(I): (II): Multiply{18−5=136=10−4​(I): (II): Subtract terms{13=136=6​ Both equations are identities, so our solution is correct.
Exercises 7 To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. In this exercise, this means that either the x-terms or the y-terms must cancel each other out. {x+y=4-3x−y=-8​(I)(II)​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {x+y=4-3x−y=-8​(II):  Add (I){x+y=4-3x−y+x+y=-8+4​ (II): Solve for x (II):  Add and subtract terms{x+y=4-2x=-4​(II):  LHS/-2=RHS/-2 {x+y=4x=2​ Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation. {x+y=4x=2​(I):  x=2{2+y=4x=2​(I):  LHS−2=RHS−2{y=2x=2​ The solution, or intersection point, of the system of equations is (2,2). To check this solution, we will substitute it back into the given system and simplify. If doing so results in true statements for every equation in the system, our solution is correct. {x+y=4-3x−y=-8​(I)(II)​(I), (II):  x=2, y=2{2+2=?4-3(2)−2=?-8​(I)(II)​(II): (-a)b=-ab{2+2=?4-6−2=?-8​(I)(II)​(I), (II):  Add and subtract terms{4=4-8=-8​(I)(II)​ Because both equations are true statements, we know our solution is correct.
Exercises 8 We will find the solution of the system of equations and then we will check the answer.Finding the solution To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the first equation by -2, the coefficients of y will be -6 and 6. Then, when we add the equations, y will cancel out, and we can solve for x. {x+3y=15x+6y=14​(I)(II)​(I):  LHS⋅(-2)=RHS⋅(-2){-2x−6y=-25x+6y=14​(I):  Add (II){-2x−6y+5x+6y=-2+145x+6y=14​(I):  Add and subtract terms{3x=125x+6y=14​(I):  LHS/3=RHS/3{x=45x+6y=14​ Now that we've found the value of x, we can substitute it into Equation (II) and solve for y. {x=45x+6y=14​(II):  x=4{x=45⋅4+6y=14​(II):  Multiply{x=420+6y=14​(II):  LHS−20=RHS−20{x=46y=-6​(II):  LHS/6=RHS/6{x=4y=-1​ The solution to the system of equations, which is the point of intersection of the two lines, is (4,-1).Checking the answer To check the answer, we will substitute 4 for x and -1 for y in both equations and simplify. {x+3y=15x+6y=14​(I)(II)​(I), (II):  x=4, y=-1{4+3(-1)=?15⋅4+6(-1)=?14​ Simplify LHS (II):  Multiply{4+3(-1)=?120+6(-1)=?14​(I), (II):  a(-b)=-a⋅b{4−3=?120−6=?14​(I), (II):  Subtract terms {1=114=14​ Since we have arrived to two identities, the solution (4,-1) is correct.
Exercises 9 To use the Elimination Method, either the x- or y-terms in both equations must cancel out when the equations are added. If we multiply the first equation by 2 and the second equation by 3, the coefficients of y will be -6 and 6. {2x−3y=-55x+2y=16​(I)(II)​(I):  LHS⋅2=RHS⋅2{4x−6y=-105x+2y=16​(II):  LHS⋅3=RHS⋅3{4x−6y=-1015x+6y=48​ Now when we add the equations, y will cancel out and we can solve for x. {4x−6y=-1015x+6y=48​(II):  Add (I){4x−6y=-1015x+6y+4x−6y=48+(-10)​(II):  Add and subtract terms{4x−6y=-1019x=38​(II):  LHS/19=RHS/19{4x−6y=-10x=2​ Substitute x=2 into Equation (I) and solve for y. {4x−6y=-10x=2​(I):  x=2{4⋅2−6y=-10x=2​(I):  Multiply{8−6y=-10x=2​(I):  LHS−8=RHS−8{-6y=-18x=2​(I):  LHS/(-6)=RHS/(-6){y=3x=2​ The solution to the system of equations, the intersection point, is (2,3).Checking the Solution We will check this solution by substituting 2 for x and 3 for y into both equations. {2x−3y=-55x+2y=16​(I)(II)​(I):  (II):  x=2, y=3{2⋅2−3⋅3=?-55⋅2+2⋅3=?16​(I):  (II):  Multiply{4−9=?-510+6=?16​(I):  (II):  Add and subtract terms{-5=-516=16​ Since our solution satisfies both equations, it is true.
Exercises 10 In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the substitution method. When solving a system of equations using substitution, there are three steps.Isolate a variable in one of the equations. Substitute the expression for that variable into the other equation and solve. Substitute this solution into one of the equations and solve for the value of the other variable. Observing the given equations, it looks like it will be simplest to isolate x in the first equation. x−y=1LHS+y=RHS+yx=1+y Now that we've isolated x, we can solve the system by substitution. {x=1+yx−y=6​(I)(II)​(II): x=1+y{x=1+y1+y−y=6​(II): Subtract term{x=1+y1​=6​ Solving this system of equations resulted in a contradiction; 1 can never be equal to 6. The lines are parallel and do not have a point of intersection. Therefore, the system does not have a solution.
Exercises 11 Let's use the elimination method to solve this system. To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {6x+2y=162x−y=2​(I)(II)​​ In its current state, this won't happen. Therefore, we need to find a common multiple between two variable like-terms in the system. If we multiply (II) by 2, the y-terms will have opposite coefficients. {6x+2y=162(2x−y)=2(2)​ ⇒ {6x+2y=164x−2y=4​​ We can see that the y-terms will eliminate each other if we add (I) to (II). {6x+2y=164x−2y=4​(II):  Add (I){6x+2y=164x−2y+6x+2y=4+16​ (II): Solve for x (II):  Add and subtract terms{6x+2y=1610x=20​(II):  LHS/10=RHS/10 {6x+2y=16x=2​ Now we can now solve for y by substituting the value of x into either equation and simplifying. {6x+2y=16x=2​(I):  x=2{6(2)+2y=16x=2​ (I): Solve for y (I):  Multiply{12+2y=16x=2​(I):  LHS−12=RHS−12{2y=4x=2​(I):  LHS/2=RHS/2 {y=2x=2​ The solution, or intersection point, of the system of equations is (2,2).
Exercises 12 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the easiest. Instead, let's use the elimination method. To use this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {3x−3y=-2-6x+6y=4​(I)(II)​​ In its current state, this won't happen. Therefore, we need to find a common multiple between two variable like-terms in the system. If we multiply (I) by 2, both the x-terms the y-terms will have opposite coefficients. {2(3x−3y)=2(-2)-6x+6y)=4​ ⇒ {6x−6y=-4-6x+6y=4​​ We can see that both the x-terms and the y-terms will eliminate each other if we add (I) to (II). {6x−6y=-4-6x+6y=4​(II):  Add (I){6x−6y=-4-6x+6y+6x−6y=4+(-4)​(II):  Add and subtract terms{6x−6y=-40=0​ Solving this system of equations resulted in an identity; 0 is always equal to itself. This means that the lines are the same and have infinitely many intersection points. Therefore the system has infinitely many solutions.
Exercises 13
Exercises 14
Exercises 15 We are told that the home team scored 6 times. They have collected 26 points from 7-point touchdowns and 3-point. If we let x represent the number of 7-pointers and y represent the number of 3-pointers, we can write the following system of equations. {7x+3y=26x+y=6​(I)(II)​ In order to solve it we will use the Substitution Method. {7x+3y=26x+y=6​(I)(II)​(II): LHS−x=RHS−x{7x+3y=26y=6−x​(I)(II)​(I): y=6−x{7x+3(6−x)=26y=6−x​(I)(II)​(I): Distribute 3{7x+18−3x=26y=6−x​(I)(II)​(I): Subtract term{4x+18=26y=6−x​(I)(II)​(I): LHS−18=RHS−18{4x=8y=6−x​(I)(II)​(I): LHS/4=RHS/4{x=2y=6−x​(I)(II)​(II): x=2{x=2y=6−2​(I)(II)​(II): Subtract term{x=2y=4​(I)(II)​ In the context of the word problem, this means that the home team scored two 7-point touchdowns and four 3-point field goals.