Exercise 9 Page 191

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 the simplest way to solve the system is to isolate x in the first equation.

2x+6y=14 & (I) 4x-8y=48 & (II)

DivEqn

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

x+3y=7 4x-8y=48

SubEqn

(I):LHS-3y=RHS-3y

x=7-3y 4x-8y=48

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

x=7-3y 4x-8y=48

Substitute

(II):x= 7-3y

x=7-3y 4( 7-3y)-8y=48

▼

(II):Solve for y

Distr

(II):Distribute 4

x=7-3y 28-12y-8y=48

SubTerm

(II):Subtract term

x=7-3y 28-20y=48

SubEqn

(II):LHS-28=RHS-28

x=7-3y -20y=20

DivEqn

(II): .LHS /(-20).=.RHS /(-20).

x=7-3y y=-1

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

x=7-3y y=-1

Substitute

(I):y= -1

x=7-3( -1) y=-1

MultNegNegOnePar

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

x=7+3 y=-1

AddTerms

(I): Add terms

x=10 y=-1

The solution, or the point of intersection, to this system of equations is the point (10,-1).