Exercise 6 Page 242

When solving a system of equations using substitution, 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.

We have been given the following system of equations. 3x-5y=13 & (I) x+4y=10 & (II) Let's start by isolating the x variable in Equation (II). x+4y=10 ⇔ x=10-4y Now, we can substitute (II) into (I) and solve for y.

3x-5y=13 & (I) x=10-4y & (II)

Substitute

(I):x= 10-4y

3( 10-4y)-5y=13 x=10-4y

Distr

(I):Distribute 3

30-12y-5y=13 x=10-4y

SubTerm

(I):Subtract term

30-17y=13 x=10-4y

SubEqn

(I):LHS-30=RHS-30

-17y=-17 x=10-4y

DivEqn

(I): .LHS /-17.=.RHS /-17.

y=1 x=10-4y

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 second equation.

y=1 & (I) x=10-4y & (II)

Substitute

(II):y= 1

y=1 x=10-4( 1)

IdPropMult

(II): Identity Property of Multiplication

y=1 x=10-4

SubTerm

(II): Subtract term

y=1 x=6

The solution, or point of intersection, to this system of equations is the point (6,1).

Checking Our Solution

To check our answer, we will substitute our solution into both equations. If doing so results in true statements, then our solution is correct.

3x-5y=13 & (I) x=10-4y & (II)

SubstituteII

x= 6, y= 1

3(6)-5(1)=13 6=10-4(1)

Multiply

(I): Multiply

18-5(1)=13 6=10-4(1)

(I), (II): Identity Property of Multiplication

18-5=13 6=10-4

(I), (II):Subtract terms

13=13 6=6

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