Exercise 8 Page 393

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. For this exercise, y is already isolated in one equation, so we can skip straight to solving!
   4x+9y=24 & (I) y=- 13x+2 & (II)

   Substitute

   (I):y= -1/3x+2

   4x+9( - 13x+2)=24 y=- 13x+2

   ▼

   (I): Simplify left-hand side

   Distr

   (I):Distribute 9

   4x+9(- 13x)+9(2)=24 y=- 13x+2

   MultPosNeg

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

   4x+(- 9( 13)x)+9(2)=24 y=- 13x+2

   AddNeg

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

   4x- 9( 13)x+9(2)=24 y=- 13x+2

   MoveLeftFacToNumOne

   (I): a* 1/b= a/b

   4x- 93x+9(2)=24 y=- 13x+2

   CalcQuot

   (I): Calculate quotient

   4x- 3x+9(2)=24 y=- 13x+2

   Multiply

   (I): Multiply

   4x-3x+18=24 y=- 13x+2

   SubTerm

   (I): Subtract term

   x+18=24 y=- 13x+2

   SubEqn

   (I):LHS-18=RHS-18

   x=6 y=- 13x+2

   Now, to find the value of y, we need to substitute x=6 into either one of the equations in the given system. Let's use the second equation.

   x=6 y=- 13x+2

   Substitute

   (II):x= 6

   x=6 y=- 13( 6)+2

   ▼

   (II): Evaluate right-hand side

   MoveRightFacToNumOne

   (II):1/b* a = a/b

   x=6 y=- 63+2

   CalcQuot

   (II):Calculate quotient

   x=6 y=-2+2

   AddTerms

   (II):Add terms

   x=6 y=0

   The solution to this system of equations is the point (6,0).