Big Ideas Math: Modeling Real Life, Grade 8
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3. Solving Systems of Linear Equations by Elimination
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Exercise 35 Page 217

Use the point-slope form to write the equations of the lines. Then, combine them into a system of linear equations and solve it using the Elimination Method.

System of Equations: y=2x & (I) y=- 13x+14 & (II)
Solution to the System: (6,12)

Practice makes perfect

We want to create a system of linear equations that represents two airplanes flying into the same airport. Let's graph the two lines using the given points on the coordinate plane and then write the equations in slope-intercept form.

We will graph two lines, a blue one to represent the path of the plane on the left and the a line for the plane on the right. Since both planes are flying towards the airport, both lines must pass through the middle point that represents the airport.

Now let's write equations of both lines in slope-intercept form. We start by finding the slopes of each line by using the Slope Formula. Let's take a look. m=y_2-y_1/x_2-x_1 In this formula, m is the slope of the line and (x_1,y_1) and (x_2,y_2) are two points that the line passes through. Keeping this in mind, let's consider the blue line.

From the graph, we can see that the blue line passes through (2,4) and (6,12). We will substitute them into the Slope Formula and evaluate to find the slope. It does not matter Notice that we can substitute the two points into the formula in any order — the answer will be the same.
m=y_2-y_1/x_2-x_1
m=12-4/6-2
m=8/4
m=2
We found that the slope of the blue line is 2. Let's do the same for the red line next. This time we will use the points (15,9) and (6,12).
m=y_2-y_1/x_2-x_1
m=12-9/6-15
m=3/-9
m=- 3/9
m=- 1/3
The slope of the red line is - 13. Before we can write the equations of the lines in slope-intercept form, we need their y-intercepts. Looking at the graph, we can see that the blue line intersects the y-axis at the point (0, 0) and the red line at the point (0, 14). y= 2x+ 0 & (I) y= - 13x+ 14 & (II) ⇕ y= 2x & (I) y= - 13x+ 14 & (II) Let's check our answer by using the Elimination Method to solve our system. Notice that the y-terms in both equations have the same coefficients. Let's subtract Equation (II) from Equation (I) to get an equation in only one variable, x.
y=2x & (I) y=- 13x+14 & (II)
y-( y)=2x-( - 13x+14) y=- 13x+14
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Solve for x
y-y=2x+ 13x-14 y=- 13x+14
0= 73x-14 y=- 13x+14
14= 73x y=- 13x+14
6=x y=- 13x+14
x=6 y=- 13x+14
Finally, let's substitute 6 for x in Equation (II) and solve for y.
x=6 & (I) y=- 13x+14 & (II)
x=6 y=- 13( 6)+14
x=6 y=-2+14
x=6 y=12
We found that the solution to the system is (6,12), which means that the two lines have a point of intersection at (6,12). This matches the given graph, so we know that the system of equations we created is correct.