Exercise 3 Page 224

When solving a system of equations using the Substitution Method, 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 = 6y-7 & (I) 4x+y=-3 & (II)

(II):x= 6y-7

x = 6y-7 4( 6y-7)+y=-3

(II): Distribute 4

x = 6y-7 24y- 28+y=-3

(II): Add terms

x = 6y-7 25y-28=-3

(II):LHS+28=RHS+28

x = 6y-7 25y=25

(II): .LHS /25.=.RHS /25.

x = 6y-7 y=1

Great! Now, to find the value of x, we need to substitute y=1 into either one of the equations in the given system. Let's use the first equation.

x = 6y-7 y=1

(I): y= 1

x=6( 1)-7 y=1

(I): Identity Property of Multiplication

x=6-7 y=1

(I): Subtract term

x=-1 y=1

The solution, or point of intersection, to this system of equations is the point (-1,1).

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= 6y-7 & (I) 4x+y=-3 & (II)

(I), (II): x= -1, y= 1

-1? =6 ( 1)-7 4 ( -1) + 1? =-3

(II): Multiply

-1 ? = 6-7 -4 + 1 ? = -3

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

-1 = -1 ✓ -3 = -3 ✓

Because both equations are true statements, we know that our solution is correct.

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