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Exercises 1 Since the coefficients of x in equation I and equation II are opposite, we will use the Elimination Method. To solve 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. {8x+3y=-9-8x+y=29​(I)(II)​​ We can see that the x-terms will be eliminated if we add (I) to (II). {8x+3y=-9-8x+y=29​(II):  Add (I){8x+3y=-9-8x+y+(8x+3y)=29+(-9)​ (II): Solve for y Remove parentheses{8x+3y=-9-8x+y+8x+3y=29−9​(II):  Simplify terms{8x+3y=-94y=20​(II):  LHS/4=RHS/4 {8x+3y=-9y=5​ Now we can now solve for x by substituting the value of y into either equation and simplifying. {8x+3y=-9y=5​(I):  y=5{8x+3(5)=-9x=-4​ (I): Solve for x (I):  Multiply{8x+15=-9y=5​(I):  LHS−15=RHS−15{8x=-24y=5​(I):  LHS/8=RHS/8 {x=-3y=5​ The solution, or intersection point, of the system of equations is (-3,5).
Exercises 2 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 solving! {21​x+y=-6y=53​x+5​(I): y=53​x+5{21​x+53​x+5=-6y=53​x+5​(I): Add terms{1011​x+5=-6y=53​x+5​(I): LHS−5=RHS−5{1011​x=-11y=53​x+5​(I):  LHS/1011​=RHS/1011​{x=-10y=53​x+5​ Great! Now, to find the value of y, we need to substitute x=-10 into either one of the equations in the given system. Let's use the second equation. {x=-10y=53​x+5​(II): x=-10{x=-10y=53​(-10)+5​(II):  Multiply{x=-10y=-6+5​(II):  Add terms{x=-10y=-1​ The solution, or point of intersection, to this system of equations is the point (-10,-1).
Exercises 3 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 solving! {y=4x+4-8x+2y=8​(I)(II)​(II): y=4x+4{y=4x+4-8x+2(4x+4)=8​(II): Multiply{y=4x+4-8x+8x+8=8​(II): Simplify terms{y=4x+48=8​ Solving this system of equations resulted in an identity; 8 is always equal to itself. Therefore, the lines are the same and have infinitely many intersection points.
Exercises 4 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, x is already isolated in one equation, so we can skip straight to solving! {x=y−11x−3y=1​(I)(II)​(II): x=y−11{x=y−11y−11−3y=1​(II): Simplify terms{x=y−11-2y−11=1​(II): LHS+11=RHS+11{x=y−11-2y=12​(II):  LHS/-2=RHS/-2{x=y−11y=-6​ Great! Now, to find the value of x, we need to substitute y=-6 into either one of the equations in the given system. Let's use the first equation. {x=y−11y=-6​(I): y=-6{x=-6−11y=-6​(I):  Subtract term{x=-17y=-6​ The solution, or point of intersection, to this system of equations is the point (-17,-6).
Exercises 5 Since neither equation has a variable with a coefficient of 1, the substitution method may not be the easiest. Instead, we'll use the elimination method. 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. {6x−4y=99x−6y=15​(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 3 and multiply (II) by -2, the x-terms will have opposite coefficients. {3(6x−4y)=3(9)-2(9x−6y)=-2(15)​ ⇒ {18x−12y=27-18x+12y=-30​​ We can see that the x-terms will eliminate each other if we add (I) to (II). {18x−12y=27-18x+12y=-30​(II):  Add (I){18x−12y=27-18x+12y+(18x−12y)=-30+(27)​Remove parentheses{18x−12y=27-18x+12y+18x−12y=-30+27​(II):  Simplify terms{12x+6y=-420=-3​ Solving this system of equations resulted in a contradiction; 0 can never be equal to -3. Therefore, the lines are parallel and do not have a point of intersection.
Exercises 6 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 solving! {y=5x−7-4x+y=-1​(I)(II)​(II): y=5x−7{y=5x−7-4x+5x−7=-1​(II): Simplify terms{y=5x−7x−7=-1​(II): LHS+7=RHS+7{y=x+5x=6​ Great! Now, to find the value of y, we need to substitute x=6 into either one of the equations in the given system. Let's use the first equation. {y=5x−7x=6​(I): x=6{y=5(6)−7x=6​(I):  Multiply{y=30−7x=6​(I):  Subtract term{y=23x=6​ The solution, or point of intersection, to this system of equations is the point (6,23).
Exercises 7 Let's plot the given points (1,2), (4,-3), (-2,8) and see where do they lie.We want points (1,2) and (4,-3) to be solutions of the system but point (-2,8) to not be a solution. Notice that both (1,2) and (4,-3) lie on the positive part of the x−axis, while (-2,8) lies on the negative part of the x-axis. We can use this observation to write the first inequality of the system. x≥0​ Let's graph it and see if it works.Since point (-2,8) doesn't belong to the solution set of the first inequality it will never be a solution to the system, so we don't have to worry about it anymore. Since the first inequality divided the coordinate plane vertically, let's write the second inequality that will divide it horizontally. In order to do so we have to look at y-coordinates of our two remaining points. They are 2 and -3, so let's take, for example, all y-values less than or equal to 4. y≤4​ This give us the following system. {x≥0y≤4​(I)(II)​​ We can graph the system to make sure it all works like we were asked to.
Exercises 8 Let's look at each of the graphs. First, we have the graph of ∣2x+1∣=∣x−7∣ which shows the points of intersection for the equations y=∣2x+1∣ and y=∣x−7∣.This graph is formed by two absolute value equations and has exactly two intersection points. Notice how the lines "bounce" back from the x-axis allowing a second point of intersection. Now, let's look at the other graph. This time we have the intersection of the lines y=4x+3 and y=-2x+9.This graph is formed by two linear equations and has exactly one intersection point. Since the lines are straight and do not "bounce" back, there cannot be a second point of intersection unless the two lines overlap completely.Similarities and Differences Similarities: In both cases we form a system of equations out of the given equation. In both cases, we are looking for points of intersection. Differences: In the first graph, there can be exactly two points of intersection while in the second this is not possible. In the second graph, there is exactly one solution.
Exercises 9 Graphing a single inequality involves two main steps.Plotting the boundary line. Shading half of the plane to show the solution set.For this exercise, we need to do this process for each of the inequalities in the system. {y>21​x+42y≤x+4​(I)(II)​​ The system's solution set will be the intersection of the shaded regions in the graphs of (I) and (II).Boundary Lines We can tell a lot of information about the boundary lines from the inequalities given in the system.Exchanging the inequality symbols for equals signs gives us the boundary line equations. Observing the inequality symbols tells us whether the inequalities are strict. Writing the equation in slope-intercept form will help us highlight the slopes m and y-intercepts b of the boundary lines.Let's find each of these key pieces of information for the inequalities in the system.InformationInequality (I)Inequality (II) Given Inequalityy>21​x+42y≤x+4 Boundary Line Equationy=21​x+42y=x+4 Solid or Dashed?> ⇒ Dashed≤ ⇒ Solid y=mx+by=21​x+4y=21​x+2 Great! With all of this information, we can plot the boundary lines.Shading the Solution Sets Before we can shade the solution set for each inequality, we need to determine on which side of the plane their solution sets lie. To do that, we will need a test point that does not lie on either boundary line.It looks like the point (0,0) would be a good test point. We will substitute this point for x and y in the given inequalities and simplify. If the substitution creates a true statement, we shade the same region the test point. Otherwise, we shade the opposite region.InformationInequality (I)Inequality (II) Given Inequalityy>21​x+42y≤x+4 Substitute (0,0)(0)>?​21​(0)+42(0)≤?​(0)+4 Simplify0≯40≤4 Shaded Regionoppositesame For Inequality (I) we will shade the region opposite test point, or above the boundary line. For Inequality (II), however, we will shade the region containing our test point, or below the boundary line.Now that we have graphed the system, we see that there is no overlapping region. This means that there are no points that are a solution to the system, only points that are solutions to each individual inequality.
Exercises 10 We have been given the following system of inequalities and asked to graph its solution set. {x+y<15x+y>4​(I)(II)​ In order to do so, we first need to identify the boundary lines.Inequality (I) Let's begin by rewriting the line in the slope-intercept form. x+y=1⇔y=-x+1 For Inequality (I), the slope is -1 and the y-intercept is 1.To choose which side of the boundary line should be shaded, we will test the point (0,0). x+y<1x=0, y=00+0<?​1Add terms0<1 Because 0<1, we will shade the region which contains the point (0,0). Note that this boundary line will be dashed because we have a strict inequality.Inequality (II) Our second boundary line will also have to be written using the slope-intercept form. 5x+y=4⇔y=-5x+4 For Inequality (II), the slope is -5 and the y-intercept is 4. To decide which region should be shaded, we can test the point (0,0). 5x+y>4x=0, y=05⋅0+0>?​4Use the Zero Product Property0+0>?​4Add terms0≯4 Because 0≯4, we will shade the region which does not contain the point (0,0). Note that this boundary line will also be dashed because we have a strict inequality.Final solution The final solution set is the overlapped region shown below.
Exercises 11 We have been given the following system of linear inequalities ⎩⎪⎨⎪⎧​y≥-32​x+1-3x+y>-2​(I)(II)​ and are asked to graph the solution sets. Let's look at graph of each individual inequality first and then combine the resulting solution sets.Inequality (I) The first thing we need to do in order to graph a linear inequality is draw the boundary line. Since our inequality is already in slope-intercept form, y=mx+b⇒y≥-32​x+1 we can draw the boundary line by considering the inequality as though it were an equation and then choosing the line to be dashed or solid based on the symbol. This time, our boundary line is y=-32​x+1. It will be solid because our symbol is ≥, meaning that points lying on the line are included in the solution set.Now we need to choose which side of the line to shade for our inequality. We can test a point in the original inequality to see if it's a solution. Let's use (0,0). y≥-32​x+1x=0, y=00≥?​-32​⋅0+1Zero Property of Multiplication0≥?​0+1Add terms0≱1 Since the point (0,0) does not satisfy the inequality, we should shade the region that does not contain that point.Inequality (II) Now we can go through the same process with the second inequality. We will start by graphing the boundary line. In order to that, let's rewrite the inequality in slope-intercept form. -3x+y>-2⇒y>3x−2 This time the line should be dashed because the symbol > indicates that points lying on the line are not included in the solution set.Let's use (0,0) again to test for our shaded region. y>3x−2x=0, y=00>?​3⋅0−2Zero Property of Multiplication0>?​0−2Subtract term0>-2 Since (0,0) satisfies the inequality, we should shade the region containing the point.Combined solution set Now we can combine the two solution sets for the individual inequalities to find the solution set for the system of inequalities.Finally, the overlapping section of the graph represents the solution set to the system of inequalities.
Exercises 12