Exercise 52 Page 384

Does either of the equations have an isolated variable in it?

(5,- 3)

Practice makes perfect

When solving a system of equations using substitution, there are three steps.

Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
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 x in Equation (I).

x+2y=- 1 & (I) 3x-5y=30 & (II)

SubEqn

(I):LHS-2y=RHS-2y

x=- 1-2y 3x-5y=30

Now that we've isolated x, we can solve the system by substitution.

x=- 1-2y & (I) 3x-5y=30 & (II)

Substitute

(II):x= - 1-2y

x=- 1-2y 3( - 1-2y)-5y=30

Distr

(II):Distribute 3

x=- 1-2y - 3-6y-5y=30

AddEqn

(II):LHS+3=RHS+3

x=- 1-2y - 6y-5y=33

SubTerms

(II):Subtract terms

x=- 1-2y - 11y=33

DivEqn

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

x=- 1-2y y=- 3

Great! Now, to find the value of x, we need to substitute y=- 3 into either one of the equations in the given system. Let's use the first equation.

x=- 1-2y y=- 3

Substitute

(I):y= - 3

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

MultNegNegOnePar

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

x=- 1+6 y=- 3

AddTerms

(I): Add terms

x=5 y=- 3

The solution, or point of intersection, to this system of equations is the point (5,- 3).

Subchapter links

Lesson Check p.381

Practice and Problem-Solving Exercises p.382-384

Standardized Test Prep p.384

Mixed Review p.384

Exercise 52 Page 384

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