Exercise 13 Page 227

When solving a system of equations using the Substitution 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.
3. Substitute this solution into one of the equations and solve for the value of the other variable.

Observing the given equations, it looks like it will be simplest to isolate x in the first equation.

2x=12 & (I) x-5y=- 29 & (II)

DivEqn

(I): .LHS /2.=.RHS /2.

x=6 x-5y=- 29

Now that we have isolated x, we can solve the system by substitution.

x=6 x-5y=- 29

Substitute

(II): x= 6

x=6 6-5y=- 29

SubEqn

(II): LHS-6=RHS-6

x=6 - 5y=- 35

DivEqn

(II): .LHS /(-5).=.RHS /(-5).

x=6 y=7

The solution, or point of intersection, to this system of equations is the point (6,7).

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.

2x=12 & (I) x-5y=- 29 & (II)

(I), (II): x= 6, y= 7

2( 6)? =12 6-5( 7)? =- 29

(I), (II): Multiply

12=12 ✓ 6-35? =- 29

SubTerm

(II): Subtract term

12=12 ✓ - 29=- 29 ✓

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