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Exercises 1 We have been asked if a system of two linear equations can have exactly two solutions. This is not possible. Let's consider why. For a system to have two solutions, the lines must intersect exactly twice causing one line to curve back around to intersect the other line a second time. However, this is not possible as the lines are straight. | |

Exercises 2 What causes a system to have no solutions? What about infinitely many solutions? The solution to a system of equations is the point of intersection. Therefore, if a system has no solutions, the functions will never ever cross paths. Also, if the system has infinitely many solutions, there are an infinite number of points where the lines touch. Let's look at the graphs of these situations.No solutions The following graph consists of two parallel lines. The defining feature of parallel lines is that they have the same slope and different y-intercepts and, therefore, will never cross paths.Infinitely many solutions The following graph consists of two identical lines. These lines have the same slope and the same y-intercepts. They will intersect at every single point on either line.Most important difference The most important difference between the two situations is the y-intercept. Systems with no solutions and with infinitely many solutions both contain functions with the same exact slope. The deciding factor is whether or not they have the same y-intercept. | |

Exercises 3 In order to find a matching graph, we begin by rewriting the system of equations in slope-intercept form. {-x+y=1x−y=1(I)(II)(I): LHS+x=RHS+x{y=x+1x−y=1(II): LHS−x=RHS−x{y=x+1-y=-x+1(II): Change signs{y=x+1y=x−1Match the graph Now we see that the graph has two parallel lines, both with a slope of 1. The y-intercepts of the lines are 1 and -1. This graph corresponds to option F.Number of solutions Parallel lines written as a system of equations will always result in an answer of "no solution." This is because parallel lines, by definition, will never intersect. They have different y-intercepts but the same slope, so they will continue on the same distance apart forever. | |

Exercises 4 In order to find a matching graph, we will begin by rewriting the system of equations in slope-intercept form. {2x−2y=4-x+y=-2(I)(II)(I): LHS−2x=RHS−2x{-2y=4−2x-x+y=-2(I): LHS/-2=RHS/-2{y=-2+x-x+y=-2(I): Rearrange factors{y=x−2-x+y=-2(II): LHS+x=RHS+x{y=x−2y=x−2Match the graph Now we can see that the graph is one line with a slope of 1. The y−intercept of this line is -2. This graph corresponds to option E.Number of solutions Because the two equations correspond to the same line, they have infinitely many intersection points. Therefore there are infinitely many solutions. | |

Exercises 5 In order to find a matching graph, we begin by rewriting the system of equations in slope-intercept form. {2x+y=4-4x−2y=-8(I)(II)(I): LHS−2x=RHS−2x{y=4−2x-4x−2y=-8(I): Rearrange factors{y=-2x+4-4x−2y=-8(II): LHS+4x=RHS+4x{y=-2x+4-2y=-8+4x(II): LHS/-2=RHS/-2{y=-2x+4y=4−2x(II): Rearrange factors{y=-2x+4y=-2x+4Match the graph Now we see that the graph is one line with a slope of -2. The y-intercept of this line is 4. This graph corresponds to option B.Number of solutions Because the two equations correspond to the same line, they have infinitely many intersection points. Therefore there are infinitely many solutions. | |

Exercises 6 In order to find a matching graph, we begin by rewriting the system of equations in slope-intercept form. {x−y=05x−2y=6(I)(II)(I): LHS+y=RHS+y{x=y5x−2y=6(I): Rearrange equation{y=x5x−2y=6(II): LHS−5x=RHS−5x{y=x-2y=6−5x(II): LHS/-2=RHS/-2⎩⎨⎧y=xy=-3+25x(II): Rearrange equation⎩⎨⎧y=xy=25x−3Match the graph This gives us two different lines: one with a slope of 1 and y−intercept 0, and the second with a slope of 25 and y−intercept -3. This graph corresponds to option C.Number of solutions As we can see, the lines intersect at one point so the system has exactly one solution. | |

Exercises 7 In order to find a matching graph, we begin by rewriting the system of equations in slope-intercept form. {-2x+4y=13x−6y=9(I)(II)(I): LHS+2x=RHS+2x{4y=1+2x3x−6y=9(I): LHS/4=RHS/4{y=41+21x3x−6y=9(I): Rearrange equation{y=21x+413x−6y=9(II): LHS−3x=RHS−3x{y=21x+41-6y=9−3x(II): LHS/-6=RHS/-6{y=21x+41y=-23+21x(II): Rearrange equation{y=21x+41y=21x−23Match the graph Now we can see that the graph has two parallel lines, both with a slope of 21. The y-intercepts of the lines are 41 and -23. This graph corresponds to option D.Number of solutions Parallel lines written as a system of equations will always result in an answer of no solution. This is because parallel lines, by definition, will never intersect. They have different y-intercepts but the same slope, so they will continue being the same distance apart forever. | |

Exercises 8 In order to find a matching graph, we begin by rewriting the system of equations in slope-intercept form. {5x+3y=17x−3y=-2(I)(II)(I): LHS−5x=RHS−5x{3y=17−5xx−3y=-2(I): LHS/3=RHS/3{y=317−35xx−3y=-2(I): Rearrange equation{y=-35x+317x−3y=-2(II): LHS−x=RHS−x{y=-35x+317-3y=-2−x(II): LHS/-3=RHS/-3{y=-35x+317y=32+31x(II): Rearrange equation{y=-35x+317y=31x+32Match the graph We have two different lines: one with a slope of -35 and y−intercept 317, and the second with a slope of 31 and y−intercept 32. This graph corresponds to option A.Number of solutions As we can see the lines intersect at one point so the system has exactly one solution. | |

Exercises 9 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. For this exercise, y is already isolated in one equation, so we can skip straight to step 2! {y=-2x−4y=2x−4(I)(II)(II): y=-2x−4{y=-2x−4-2x−4=2x−4(II): LHS+2x=RHS+2x{y=-2x−4-4=4x−4(II): LHS+4=RHS+4{y=-2x−40=4x(II): LHS/4=RHS/4{y=-2x−40=x(II): Rearrange equation{y=-2x−4x=0 Now, to find the value of y, we need to substitute x=0 into either one of the equations in the given system. Let's use the first equation. {y=-2x−4x=0(I): x=0{y=-2(0)−4x=0(I): Zero Property of Multiplication{y=-4x=0 The solution to this system of equations is the point of intersection (0,-4). | |

Exercises 10 Because both of the given equations has a coefficient of 1 for the y variable, we can solve this system using elimination by subtraction. {y=-6x−8y=-6x+8(I)(II)(I): Subtract (II){y−y=-6x−8−(-6x+8)y=-6x+8(I): -(b−a)=a−b{y−y=-6x−8−8+6xy=-6x+8(I): Add and subtract terms{0≠-16y=-6x+8 We have reached a contradiction, 0 can never equal -16. This contradiction means that there are no solutions to this system of equations. Alternative solution Another way of thinking If you notice, both equations are in slope-intercept form and have slopes of -6. When two equations have the same slope but different y-intercepts, they are parallel. Parallel lines will never intersect and, therefore, the system will have no solutions. | |

Exercises 11 Since the coefficients of the variables are additive inverses, the Elimination Method seems to be the best approach. To use this method to solve the system of equations, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {3x−y=6-3x+y=-6(I)(II) We can see that both the x-terms and y-terms will eliminate each other if we add (I) to (II). {3x−y=6-3x+y=-6(II): Add (I){3x−y=6-3x+y+3x−y=-6+6(II): Add and subtract terms{3x−y=60=0 Solving this system of equations resulted in an identity; 0 is always equal to itself. The lines are the same and have infinitely many intersection points. Therefore we know that the system has infinitely many solutions. | |

Exercises 12 Since the coefficients of the variables are additive inverses the Elimination Method seems to be the best approach. To use this method to solve the system of equations, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {-x+2y=7x−2y=7(I)(II) We can see that both the x-terms and y-terms will eliminate each other if we add (I) to (II). {-x+2y=7x−2y=7(II): Add (I){-x+2y=7x−2y+(-x+2y)=7+7(II): a+(-b)=a−b{-x+2y=7x−2y−x+2y=7+7(II): Add and subtract terms{-x−2y=70≠14 Solving this system of equations resulted in a contradiction; 0 can never be equal to 14. The lines are parallel and do not have a point of intersection. Therefore, we know that there is no solution to this system of equations. | |

Exercises 13 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the best approach. Instead, we'll use the Elimination Method. To use this method to solve the system, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {4x+4y=-8-2x−2y=4(I)(II) In its current state, this won't happen. However if we multiply (II) by 2, the x-terms and the y-terms will have opposite coefficients. {4x+4y=-82(-2x−2y)=2(4) ⇒ {4x+4y=-8-4x−4y=8 We can see that both the x-terms and the y-terms will eliminate each other if we add (I) to (II). {4x+4y=-8-4x−4y=8(II): Add (I){4x+4y=-8-4x−4y+4x+4y=-8+8(II): Add and subtract terms{4x+4y=-80=0 Solving this system of equations resulted in an identity; 0 is always equal to itself. The lines are the same and have infinitely many intersection points. Therefore, we know that the system of equations has infinitely many solutions. | |

Exercises 14 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 y in the second equation. -3x+y=4LHS+3x=RHS+3xy=4+3x Now that we've isolated y, we can solve the system by substitution. {15x−5y=-20y=4+3x(I)(II)(I): y=4+3x{15x−5(4+3x)=-20y=4+3x(I): Distribute -5{15x−20−15x=-20y=4+3x(II): Subtract term{-20=-20y=4+3x Solving this system of equations resulted in an identity; -20 is always equal to itself. The lines are the same and have infinitely many intersection points. Therefore, we know that the system of equations has infinitely many solutions. | |

Exercises 15 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the best approach. Instead, we'll use the Elimination Method. To use this method to solve a system, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {9x−15y=246x−10y=-16(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 divide (I) by -3 and divide (II) by 2, both the x-terms and y-terms will have opposite coefficients. {(9x−15y)÷(-3)=24÷(-3)(6x−10y)÷2=(-16)÷2 ⇒ {-3x+5y=-83x−5y=-8 We can see that both the x-terms and the y-terms will eliminate each other if we add (I) to (II). {-3x+5y=-83x−5y=-8(II): Add (I){-3x+5y=-83x−5y+(-3x+5y)=-8+(-8)(II): a+(-b)=a−b{-3x+5y=-83x−5y−3x+5y=-8−8(II): Add and subtract terms{-3x+5y=-80≠-16 Solving this system of equations resulted in a contradiction; 0 can never be equal to -16. This means the lines are parallel and do not have a point of intersection. Therefore, we know that this system of equations has no solution. | |

Exercises 16 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the best approach. Instead, we'll use the Elimination Method. To use this method to solve a system, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. {3x−2y=-54x+5y=47(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 5 and multiply (II) by 2, the y-terms will have opposite coefficients. {5(3x−2y)=5(-5)2(4x+5y)=2(47) ⇒ {15x−10y=-258x+10y=94 We can see that the y-terms will eliminate each other if we add (I) to (II). {15x−10y=-258x+10y=94(II): Add (I){15x−10y=-258x+10y+15x−10y=94+(-25) (II): Solve for x (II): Add and subtract terms{15x−10y=-2523x=69(II): LHS/23=RHS/23 {15x−10y=-25x=3 Now we can now solve for y by substituting the value of x into either equation and simplifying. Let's use the first equation. {15x−10y=-25x=3(I): x=3{15⋅3−10y=-25x=3 (I): Solve for y (I): Multiply{45−10y=-25x=3(I): LHS−45=RHS−45{-10y=-70x=3(I): LHS/-10=RHS/-10 {y=7x=3 The solution of the system of equations is the point of intersection (3,7). | |

Exercises 17 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b y=7x+13y=7x+137(0,13) -21x+3y=39y=7x+137(0,13) Comparing the slopes, we see that they are equal, so the lines are either parallel or the same. Looking at the y-intercepts, we can see that the point at which both lines cross the y-axis is 13. This means that the lines are the same and the system has infinitely many solutions. | |

Exercises 18 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b y=-6x−2y=-6x+(-2)-6(0,-2) 12x+2y=-6y=-6x+(-3)-6(0,-3) Comparing the slopes, we see that they are equal, so the lines are either parallel or the same. Looking at the y-intercepts, we can tell the lines are different because the point at which each line crosses the y-axis is different. This means that the lines are parallel and the system has no solution. | |

Exercises 19 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b 4x+3y=27y=-34x+9-34(0,9) 4x−3y=-27y=34x+934(0,9) Comparing the slopes, we see that they are not equal. Therefore we know that the lines are neither parallel nor the same, and intersect at one point. This means that the system has one solution. | |

Exercises 20 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b -7x+7y=1y=1x+711(0,71) 2x−2y=-18y=1x+91(0,9) Comparing the slopes, we see that they are equal, so the lines are either parallel or the same. Looking at the y-intercepts, we can tell the lines are different because the point at which each line crosses the y-axis is different. This means that the lines are parallel and the system has no solution. | |

Exercises 21 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b -18x+6y=24y=3x+43(0,4) 3x−y=-2y=3x+23(0,2) Comparing the slopes, we see that they are equal, so the lines are either parallel or the same. Looking at the y-intercepts, we can tell the lines are different because the point at which each line crosses the y-axis is different. This means that the lines are parallel and the system has no solution. | |

Exercises 22 An alternative method to determining the number of solutions to a system of equations by graphing is to compare the slope and y-intercept of the equations. y=mx+b To do this, use the slope-intercept form of each equation, where m is the slope and the point (0,b) is the y-intercept. There are three possibilities when comparing two linear equations in a system.Slopey-interceptGraph DescriptionNumber of Solutions m1≠m2irrelevantintersecting linesone solution m1=m2b1≠b2parallel linesno solution m1=m2b1=b2same lineinfinitely many Let's write the equations in the given system in slope-intercept form, highlighting the m and b values.Given EquationSlope-Intercept FormSlope my-intercept b 2x−2y=16y=1x+(-8)1(0,-8) 3x−6y=30y=21x+(-5)21(0,-5) Comparing the slopes, we see that they are not equal. Therefore we know the lines are neither parallel nor the same, and intersect at one point. This means that the system has one solution. | |

Exercises 23 In order to see what went wrong, let's first rewrite the equations in our system into slope-intercept form. {-4x+y=44x+y=12(I)(II)(I): LHS+4x=RHS+4x{y=4x+44x+y=12(II): LHS−4x=RHS−4x{y=4x+4y=-4x+12 These lines have different slopes, so they must intersect and have one solution. We do not see the intersection point on the given diagram because the axes are not long enough. Let's extend the given graph.W see that the lines intersect at (1,8). | |

Exercises 24 To see why this answer is incorrect, let's graph the equations. on the same coordinate plane. Note that both lines are written in slope-intercept form. Slope-intercept form:Equation I:Equation II: y=mx+b y=3x+(-8) y=3x+(-12) We will plot the y-intercepts and then use the slopes to find a second point on each line. Then, we will connect the points using a straight edge.We see that the lines are parallel and therefore the system has no solution. In fact, if two lines have the same slope but different y-intercept, they are parallel lines and the system has no solution. The error was thinking the lines were the same line just because they have the same slope. | |

Exercises 25 If we let price of almonds be a and the price of dried fruits be f, then the algebraic expressions that represent the contents of each type of bag are: Small bag: Large bag: 3f+4a4.5f+6a. We can equate these expressions with their total costs, $6 for the small bag and $9 for the large bag, to create our system of equations. {3f+4a=64.5f+6a=9 In order to solve this system, we should multiply the first equation by 3 and the second equation by 2 so that both equations have 9f. Then we can solve using elimination by subtraction. {3f+4a=64.5f+6a=9(I)(II)(I): LHS⋅3=RHS⋅3{9f+12a=184.5f+6a=9(II): LHS⋅2=RHS⋅2{9f+12a=189f+12a=18(I): Subtract (II){9f+12a−(9f+12a)=18−189f+12a=18(I): Remove parentheses and change signs{9f+12a−9f−12a=18−189f+12a=18(I): Subtract terms{0=09f+12a=18 In solving, we have reached an identity, 0 will always equal 0. This means that there are infinitely many solutions. | |

Exercises 26 We want to write a system of linear equations to represent the situation described. Let's start by recalling the slope-intercept form of a line. In this form, m represents the slope and b the y-intercept. y=mx+b Let x be the time, in hours, that has passed after the situation is observed. Let y be the position of the second canoe in the moment the situation is observed. Since we are told Team A is 2 miles ahead of Team B, we can write the y-intercepts and start defining our system. Team A:Team B: y=max+2 y=mbx+0⇓{y=max+2y=mbx The slopes will be determined by the speed. Both canoes travel six miles in an hour, and therefore their slope is 6. {y=6x+2(A)y=6x+2(B) Paying close attention to the system above, we see that the slope is the same for both equations. This means the lines are parallel. Therefore, they do not intersect and the system has no solution. Team B will never catch up to Team A. | |

Exercises 27 If we let the cost of each coach ticket be c and the cost of each business class ticket be b, then we can write the following system of equations: {150c+80b=22860170c+100b=27280 These equations tell us that the number of each kind of ticket sold, multiplied by the cost of each kind of ticket, added together will give the amount of total money collected. We can solve this system to find the cost of one coach ticket. Before we can solve the system, we need to multiply the equations to get equal coefficients. {150c+80b=22860170c+100b=27280(I)(II)(I): LHS⋅5=RHS⋅5{750c+400b=114300170c+100b=27280(I): LHS⋅4=RHS⋅4{750c+400b=114300680c+400b=109120 Now that our coefficients for b are equal, we can solve for c using elimination by subtraction. {750c+400b=114300680c+400b=109120(I): Subtract (II){750c+400b−(680c+400b)=114300−109120680c+400b=109120(I): Remove parentheses and change signs{750c+400b−680c−400b=114300−109120680c+400b=109120(I): Subtract terms{70c=5180680c+400b=109120(I): LHS/70=RHS/70{c=74680c+400b=109120 One coach ticket costs $74. | |

Exercises 28 We want to write a system of three linear equations in two variables that satisfies certain conditions. Let's pay close attention to those conditions.There exists one solution for each combination of two equations. This tells us that none of them can be parallel. There is no solution for all three equations. This tells us that they cannot all intersect each other at a single point.It might be easier to draw the lines on a graph first and then write the system based on the graph. Note that there are infinitely many solutions to this exercise. Here we will show only one possible solution.We see that any two of the lines intersect at exactly one point, but the entire system has no solution. Therefore, the three lines graphed above form a system which satisfies the required conditions. ⎩⎪⎨⎪⎧y=5y=xy=-x+5 | |

Exercises 29 A system of linear equations can either have infinitely many solutions, no solutions, or one solution. Let's think about what leads us to having these different results.Infinitely many solutions: The functions are the exact same line (they have the same slope and y-intercept). There are infinitely many solutions because the lines will intersect at every single possible point. No solutions: The functions are parallel lines (they have the same slope but different y-intercepts). There are no solutions because the lines will never intersect. One solution: The functions are nonparallel lines (they have different slopes). These lines will always intersect only once due to the continuous nature of linear functions.The equations we have been asked to consider have slopes of 2 and -31. These are going to be nonparallel lines and will, therefore, intersect one time and have only one solution. | |

Exercises 30 aWhen looking at the graph, we can tell when Team C's runner passed Team B's runner by looking at the point of intersection. Once the green line has higher y values than the red line, Team C has ran a farther distance and is ahead of Team B.This seems to happen around the point (8,40). Since the y values represent the distance, we can say that Team B was passed after around 40 meters.bIf the race was longer, and the runners maintained the same speed, Team C's runner definitely could have passed Team A's runner. To visualize this, let's extend our graph.cRecall that the speed is the quotient between distance and time. In this case, the time is represented by the horizontal axis and the distance by the vertical axis. Moreover, the slope of a line is calculated by doing rise over run. With this in mind, we conclude that the slope of the lines represent the speed of the runners. speed=timedistanceandslope=runrise⇓speed=slope We are told that Team A's runner and Team B's runner both have a speed of 7.8 meters per second. Therefore, the lines which represent each runner have the same slope. This means the lines are parallel, and they never intersect. Therefore, if the race was longer, Team B's runner could never have passed Team A's runner. | |

Exercises 31 aIn order for a system of linear equations to have infinitely many solutions, the lines need to be identical. They should have the same slope and y-intercept. We have been given the following system: {y=ax+4y=bx−2. Because this system has one equation with a y-intercept at 4 and another with a y-intercept at -2, there will never be a case such that there are infinitely many solutions.bWhen a system of linear equations has no solutions, the lines in the system are parallel. In the given system, {y=ax+4y=bx−2, the lines will be parallel anytime a=b. Therefore, the system will sometimes have no solution.cA system of linear equations will have one solution any time the lines have different slopes. If a<b, then a≠b and the slopes cannot be the same. Therefore, in the given system, {y=ax+4y=bx−2, when a<b there will always be one solution. | |

Exercises 32 If we let a be the cost of one admission and s be the cost of skate rentals then, using the receipts, we can write the following system of equations: {3a+2s=3815a+10s=190 Let's see if we can solve this system and find the cost of one admission and one skate rental. First we should multiply the first equation by 5 so that we can have equal coefficients. {3a+2s=3815a+10s=190(I)(II)(I): LHS⋅5=RHS⋅5{15a+10s=19015a+10s=190 We can see that our system of equations consists of two identical lines. This means that there exist infinitely many solutions. With infinitely many solutions there is no way to tell the exact cost of the admissions and skate rentals. | |

Exercises 33 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d)Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. ∣2x+6∣=∣x∣2x+6≥0: 2x+6=(x)2x+6<0: 2x+6=-(x)(I)(II)2x+6=(x)2x+6=-(x)(I)(II) Solve for x (II): Distribute -12x+6=x2x+6=-x(I): LHS−x=RHS−xx+6=02x+6=-x(I): LHS−6=RHS−6x=-62x+6=-x(II): LHS+x=RHS+xx=-63x+6=0(II): LHS−6=RHS−6x=-63x=-6(II): LHS/3=RHS/3 x=-6x=-2 We got two solutions x=-6 and x=-2. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. We will start with x=-6. ∣2x+6∣=∣x∣x=-6∣2(-6)+6∣=?∣-6∣ Simplify a(-b)=-a⋅b∣-12+6∣=?∣-6∣Add terms∣-6∣=?∣-6∣∣-6∣=6 6=6 Therefore x=-6 is not extraneous. Now, let's check x=-2. ∣2x+6∣=∣x∣x=-2∣2(-2)+6∣=?∣-2∣ Simplify a(-b)=-a⋅b∣-4+6∣=?∣-2∣\Addterms∣-2∣=?∣-2∣∣-2∣=2 2=2 So x=-2 is not extraneous either. | |

Exercises 34 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d)Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. ∣3x−45∣=∣12x∣3x−45≥0: 3x−45=(12x)3x−45<0: 3x−45=-(12x)(I)(II)3x−45=(12x)3x−45=-(12x)(I)(II) Solve for x (II): Distribute -13x−45=12x3x−45=-12x(I): LHS−3x=RHS−3x-45=9x3x−45=-12x(I): LHS/9=RHS/9-5=x3x−45=-12x(I): Rearrange equationx=-53x−45=-12x(II): LHS+12x=RHS+12xx=-515x−45=0(II): LHS+45=RHS+45x=-515x=45(II): LHS/15=RHS/15 x=-5x=3 We got two solutions x=-5 and x=3. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. We will start with x=-5. ∣3x−45∣=∣12x∣x=-5∣3(-5)−45∣=?∣12(-5)∣ Simplify a(-b)=-a⋅b∣-15−45∣=?∣-60∣Subtract term∣-60∣=?∣-60∣∣-60∣=60 60=60 Therefore x=-5 is not extraneous. Now, let's check x=3. ∣3x−45∣=∣12x∣x=3∣3(3)−45∣=?∣12(3)∣ Simplify Multiply∣9−45∣=?∣36∣Subtract term∣-36∣=?∣36∣∣-36∣=36 36=36 So x=3 is not extraneous either. | |

Exercises 35 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d)Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. ∣x−7∣=∣2x−8∣x−7≥0: x−7=(2x−8)x−7<0: x−7=-(2x−8)(I)(II)x−7=(2x−8)x−7=-(2x−8)(I)(II) Solve for x (II): Distribute -1x−7=(2x−8)x−7=-2x+8(I): LHS−x=RHS−x-7=x−8x−7=-2x+8(I): LHS+8=RHS+81=xx−7=-2x+8(I): Rearrange equationx=1x−7=-2x+8(II): LHS+2x=RHS+2xx=13x−7=8(II): LHS+7=RHS+7x=13x=15(II): LHS/3=RHS/3 x=1x=5 We found two solutions x=1 and x=5. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. We will start with x=1. ∣x−7∣=∣2x−8∣x=1∣1−7∣=?∣2(1)−8∣ Simplify Multiply∣1−7∣=?∣2−8∣Subtract terms∣-6∣=?∣-6∣∣-6∣=6 6=6 Therefore x=1 is not extraneous. Now, let's check x=5. ∣x−7∣=∣2x−8∣x=5∣5−7∣=?∣2(5)−8∣ Simplify Multiply∣5−7∣=?∣10−8∣Subtract terms∣-2∣=?∣2∣∣-2∣=2 2=2 So x=5 is not extraneous either. | |

Exercises 36 When solving an equation involving absolute value expressions, we should consider what would happen if we removed the absolute value symbols. Let's look at an example equation. ∣ax+b∣=∣cx+d∣ Although we can make 4 statements about this equation, there are actually only two possible cases to consider.StatementResult Both absolute values are positive.ax+b=cx+d Both absolute values are negative.-(ax+b)=-(cx+d) Only the left-hand side is negative.-(ax+b)=cx+d Only the right-hand side is negative.ax+b=-(cx+d)Because of the Properties of Equality, when the absolute values of two expressions are equal, either the expressions are equal or the opposites of the expressions are equal. Now let's consider these two cases for the given equation. ∣2x+1∣=∣3x−1∣2x+1≥0: 2x+1=(3x−1)2x+1<0: 2x+1=-(3x−1)(I)(II)2x+1=(3x−1)2x+1=-(3x−1)(I)(II) Solve for x (II): Distribute -12x+1=3x−12x+1=-3x+1(I): LHS−2x=RHS−2x1=x−12x+1=-3x+1(I): LHS+1=RHS+12=x2x+1=-3x+1(I): Rearrange equationx=22x+1=-3x+1(II): LHS+3x=RHS+3xx=25x+1=1(II): LHS−1=RHS−1x=25x=0(II): LHS/5=RHS/5 x=2x=0 We found two solutions x=2 and x=0. After solving an absolute value equation, it is necessary to check for extraneous solutions. To do this, we substitute the found solutions into the given equation and determine if a true statement is made. We will start with x=2. ∣2x+1∣=∣3x−1∣x=2∣2(2)+1∣=?∣3(2)−1∣ Simplify Multiply∣4+1∣=?∣6−1∣Add and subtract terms∣5∣=?∣5∣∣5∣=5 5=5 Therefore x=2 is not extraneous. Now, let's check x=0. ∣2x+1∣=∣3x−1∣x=0∣2(0)+1∣=?∣3(0)−1∣ Simplify Zero Property of Multiplication∣1∣=?∣-1∣∣-1∣=1 1=1 So x=0 is not extraneous either. |

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##### Other subchapters in Solving Systems of Linear Equations

- Maintaining Mathematical Proficiency
- Mathematical Practices
- Solving Systems of Linear Equations by Graphing
- Solving Systems of Linear Equations by Substitution
- Solving Systems of Linear Equations by Elimination
- Quiz
- Solving Equations by Graphing
- Graphing Linear Inequalities in Two Variables
- Systems of Linear Inequalities
- Chapter Review
- Chapter Test
- Cumulative Assessment