Core Connections: Course 3
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2. Section 8.2
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Exercise 93 Page 368

Practice makes perfect

We want to find the coordinates of the point of intersection for the following system of equations. y= & 2x-3 y= & 4x+1 We can do this by solving the system of equations. system, we are given two different expressions for the value of y. This means that we can find the solution using the Equal Values Method. When solving a system of equations using this method, there are three steps.

  1. Isolate a variable in one of the equations.
  2. Substitute the expression for that variable into the other equation and solve for the remaining variable.
  3. Substitute this solution into one of the equations and solve for the value of the first variable.
For this system, y is already isolated in both equations, so we can skip straight to solving!
y=2x-3 & (I) y=4x+1 & (II)
y=2x-3 2x-3=4x+1
y=2x-3 2x-3-2x=4x+1-2x
y=2x-3 - 3=2x+1
y=2x-3 - 3-1=2x+1-1
y=2x-3 - 4=2x
y=2x-3 - 42= 2x2
y=2x-3 - 42=x
y=2x-3 - 42=x
y=2x-3 - 2=x
y=2x-3 x=- 2
Great! Now, to find the value of y, we need to substitute x=- 2 into either one of the equations in the given system. Let's use the first equation.
y=2x-3 x=- 2
y=2( - 2)-3 x=- 2
y=- 4-3 x=- 2
y=- 7 x=- 2
The solution, or point of intersection, to this system of equations is the point (- 2,- 7).

Checking Our Answer

Let's Check!
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=2x-3 & (I) y=4x+1 & (II)

(I), (II): x= - 2, y= - 7

- 7? =2( - 2)-3 - 7? =4( - 2)+1

(I), (II): a(- b)=- a * b

- 7? =- 4-3 - 7? =- 8+1

(I), (II): Add and subtract terms

- 7=- 7 âś“ - 7=- 7 âś“
Because both equations are true statements, we know that our solution is correct.
We are asked to find the coordinates of the point of intersection for the following system of equations. y = & 2x-5 y = & - 4x-2 Let's solve the system of equations. Like in Part A, we will use the Equal Values Method. Since y is already isolated in both equations, so we can set the expressions equal to each other and solve for x.
y=2x-5 & (I) y=- 4x-2 & (II)
y=2x-5 2x-5=- 4x-2
y=2x-5 6x-5=- 2
y=2x-5 6x=3
y=2x-5 6x6= 36
y=2x-5 x= 36
y=2x-5 x= 3Ă· 36Ă· 3
y=2x-5 x= 12
Next, let's find the value of y by substituting x= 12 into the first equation.
y=2x-5 x= 12
y=2( 12)-5 x= 12
y=1-5 x= 12
y=- 4 x= 12
The solution, or point of intersection, to the system of equations is the point ( 12,- 4).

Checking Our Answer

Let's Check!
We can check our answer by substituting our solution into both equations. If doing so results in true statements, then our solution is correct.
y=2x-5 & (I) y=- 4x-2 & (II)

(I), (II): x= 1/2, y= - 4

- 4? =2( 12)-5 - 4? =- 4( 12)-2

(I), (II): a* 1/b= a/b

- 4? = 22-5 - 4? =- 42-2

(I), (II): Calculate quotient

- 4? =1-5 - 4? =- 2-2

(I), (II): Subtract terms

- 4=- 4 âś“ - 4=- 4 âś“
Since the left-hand side is equal to the right-hand side, our solution is correct.