Exercise 1 Page 145

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.

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

3x+5y=13 & (I) 2x+y=4 & (II)

SubEqn

(II): LHS-2x=RHS-2x

3x+5y=13 y=4-2x

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

3x+5y=13 & (I) y=4-2x & (II)

Substitute

(I): y= 4-2x

3x+5( 4-2x)=13 y=4-2x

▼

(I):Solve for x

Distr

(I): Distribute 5

3x+20-10x=13 y=4-2x

SubTerm

(I): Subtract term

-7x+20=13 y=4-2x

SubEqn

(I): LHS-20=RHS-20

-7x=-7 y=4-2x

DivEqn

(I): .LHS /(-7).=.RHS /(-7).

x=1 y=4-2x

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

x=1 y=4-2x

Substitute

(II):x= 1

x=1 y=4-2( 1)

IdPropMult

(II): Identity Property of Multiplication

x=1 y=4-2

SubTerm

(II): Subtract term

x=1 y=2

The solution, or point of intersection, to this system of equations is the point (1,2).