Exercise 10 Page 228

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.

Since the expression equal to y in (II) is simpler, let's use that for our initial substitution.

y=-3x-7 & (I) y=x+9 & (II)

Substitute

(I):y= x+9

x+9=-3x-7 y=x+9

▼

(I): Solve for x

SubEqn

(I): LHS-9=RHS-9

x+9-9=-3x-7-9 y=x+9

SubTerm

(I): Subtract term

x=-3x-16 y=x+9

AddEqn

(I): LHS+3x=RHS+3x

x+3x=-3x-16+3x y=x+9

AddTerms

(I): Add terms

4x=- 16 y=x+9

DivEqn

(II): .LHS /4.=.RHS /4.

4x4= - 164 y=x+9

CrossCommonFac

(I): Cross out common factors

4x4= - 164 y=x+9

SimpQuot

(I): Simplify quotient

x= - 164 y=x+9

CalcQuot

(I): Calculate quotient

x=- 4 y=x+9

Great! Now, to find the value of y, we need to substitute x=- 4 into either one of the equations in the given system. Let's use the first equation.

x=- 4 & (I) y=x+9 & (II)

Substitute

(I):x= - 4

x=- 4 y= - 4+9

AddTerms

(I): Add terms

x=- 4 y=5

The solution, or point of intersection, to this system of equations is the point (- 4,5).

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.

y=-3x-7 & (I) y=x+9 & (II)

(I), (II): x= - 4, y= 5

5? =-3( - 4)-7 5? = - 4+ 9

MultNegNegOnePar

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

5? =3* 4-7 5? =- 4+9

Multiply

(I): Multiply

5? =12-7 5? =- 4+9

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

5=5 5=5

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