Exercise 4 Page 230

In this system of equations, at least one of the variables has a coefficient of 1. Therefore, we will approach its solution with the Substitution Method. - 2 x + y +3 = 0 & (I) 3x+4y = - 1 & (II) 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 y in the first equation.

- 2 x + y +3 = 0 & (I) 3x+4y = - 1 & (II)

AddEqn

(I): LHS+2x=RHS+2x

y +3 = 2x 3x+4y = - 1

SubEqn

(I): LHS-3=RHS-3

y = 2x-3 3x+4y = - 1

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

y = 2x-3 3x+4y = - 1

Substitute

(II):y= 2x-3

y = 2x-3 3x+4( 2x-3) = - 1

Distr

(II): Distribute 4

y = 2x-3 3x+8x-12 = - 1

AddTerms

(II):Add terms

y = 2x-3 11x-12 = - 1

AddEqn

(II):LHS+12=RHS+12

y = 2x-3 11x = 11

DivEqn

(II): .LHS /11.=.RHS /11.

y = 2x-3 x = 1

Great! Now to find the value of y, we need to substitute x=1 into the first equation.

y = 2x-3 x = 1

Substitute

(I):x= 1

y = 2( 1)-3 x = 1

Multiply

(I): Multiply

y = 2-3 x = 1

SubTerms

(I): Subtract terms

y=- 1 x=1

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