Core Connections Geometry, 2013
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Core Connections Geometry, 2013 View details
1. Section 9.1
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Exercise 11 Page 534

Practice makes perfect
a In this system of equations, at least one of the variables has a coefficient of 1. Let's use the Substitution Method to solve the system. When solving a system using this method, we follow 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 value of the second equation.
  3. Substitute this solution into one of the equations and solve for the value of the first variable.
For this exercise, x is already isolated in one equation, so we can skip straight to solving!
3x-y=14 & (I) x=2y+8 & (II)
3( 2y+8)-y=14 x=2y+8
6y+24-y=14 x=2y+8
5y+24=14 x=2y+8
5y=-10 x=2y+8
y=-2 x=2y+8
Great! Now, to find the value of x, we will substitute y=-2 into either one of the equations in the given system. Let's use the second equation.
y=-2 x=2y+8
y=-2 x=2( -2)+8
y=-2 x=-4+8
y=-2 x=4
The solution, or point of intersection, of this system of equations is the point (4,-2).

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. Let's do it!
3x-y=14 & (I) x=2y+8 & (II)

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

3( 4)-( -2)? =14 4? =2( -2)+8
3(4)+2? =14 4? =2(-2)+8

(I), (II): Multiply

12+2? =14 4? =-4+8

(I), (II): Add terms

14=14 âś“ 4=4 âś“
Because both equations are true statements, we know that our solution is correct.
b At least one of the variables in this system of equations has a coefficient of 1, so let's use the Substitution Method to solve it. When solving using this method, we follow 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 value of the second equation.
  3. Substitute this solution into one of the equations and solve for the value of the first variable.
Since the expression equal to x in (II) is simpler, let's use that for our initial substitution.
x=2y+2 & (I) x=- y-10 & (II)
- y-10=2y+2 x=- y-10
- y=2y+12 x=- y-10
- 3y=12 x=- y-10
y=-4 x=- y-10
Great! Now to find the value of x, we need to substitute y=-4 into either one of the equations in the given system. Let's use the second equation.
y=-4 x=- y-10
y=-4 x=- ( -4)-10
y=-4 x=4-10
y=-4 x=-6
The solution, or point of intersection, to this system of equations is the point (-6,-4).

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.
x=2y+2 & (I) x=- y-10 & (II)

(I), (II): x= -6, y= -4

-6? =2( -4)+2 -6? =- ( -4)-10
-6? =2(-4)+2 -6? =4-10
-6? =-8+2 -6? =4-10

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

-6=-6 âś“ -6 = -6 âś“
Because both equations are true statements, we know that our solution is correct.
c Even though y in the second equation has a coefficient of 1, the Substitution Method may not be the easiest. Instead, let's try using the Elimination Method. To solve a system using this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other. This means that either the x-terms or the y-terms must cancel each other out.
16 x- y=-4 & (I) 2 x+ y=13 & (II) We can see that the y-terms will eliminate each other if we add Equation (I) to Equation (II).
16x-y=-4 2x+y=13
16x-y=-4 2x+y+( 16x-y)=13+( -4)
â–Ľ
(II):Solve for x
16x-y=-4 18x=9
16x-y=-4 x= 12
Now we can solve for y by substituting the value of x into either equation and simplifying. Let's do it!
16x-y=-4 x= 12
16( 12)-y=-4 x= 12
â–Ľ
(I):Solve for y
8-y=-4 x= 12
8=y-4 x= 12
12=y x= 12
y=12 x= 12
The solution, or point of intersection, of the system of equations is ( 12,12).

Checking Our Answer

To check our answer, 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.
16x-y=-4 & (I) 2x+y=13 & (II)

(I), (II): x= 1/2, y= 12

16( 12)- 12? =-4 2( 12)+ 12? =13

(I), (II): Multiply

8-12? =-4 1+12? =13

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

-4 = -4 âś“ 13=13 âś“