Exercise 5 Page 274

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 x in the first equation.
   3.25x-1.5y=1.25 & (I) 13x-6y=10 & (II)

   AddEqn

   (I): LHS+1.5y=RHS+1.5y

   3.25x=1.25+1.5y 13x-6y=10

   WriteAsFrac

   (I): Write as a fraction

   134x= 54+ 32y 13x-6y=10

   MultEqn

   (I): LHS * 4/13=RHS* 4/13

   ( 134x) 413=( 54+ 32y) 413 13x-6y=10

   ▼

   Simplify left-hand side

   CommutativePropMult

   (I): Commutative Property of Multiplication

   ( 134) 413x=( 54+ 32y) 413 13x-6y=10

   MultFracByInverse

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

   1x=( 54+ 32y) 413 13x-6y=10

   IdPropMult

   (I): Identity Property of Multiplication

   x=( 54+ 32y) 413 13x-6y=10

   ▼

   (I): Simplify right-hand side

   CommutativePropMult

   (I): Commutative Property of Multiplication

   x= 413( 54+ 32y) 13x-6y=10

   MoveNumRight

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

   x= 113[4( 54+ 32y)] 13x-6y=10

   Distr

   (I): Distribute 4

   x= 113[4( 54)+4( 32y)] 13x-6y=10

   DenomMultFracToNumber

   (I): 4 * a/4= a

   x= 113[5+4( 32y)] 13x-6y=10

   MoveLeftFacToNum

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

   x= 113[5+ 4(3)2y] 13x-6y=10

   Multiply

   (I): Multiply

   x= 113(5+ 122y) 13x-6y=10

   CalcQuot

   (I): Calculate quotient

   x= 113(5+6y) 13x-6y=10

   Great! Now, to find the value of y, we need to substitute x= 113(5+6y) into the second equation.

   x= 113(5+6y) 13x-6y=10

   Substitute

   (II):x= 1/13(5+6y)

   x= 113(5+6y) 13* 113(5+6y)-6y=10

   (II): a * 1/a=1

   x= 113(5+6y) 1(5+6y)-6y=10

   ▼

   (II): Simplify left-hand side

   Distr

   (II): Distribute 1

   x= 113(5+6y) 1* 5+1* 6y-6y=10

   IdPropMult

   (II): Identity Property of Multiplication

   x= 113(5+6y) 5+6y-6y=10

   SubTerm

   (II): Subtract term

   x= 113(5+6y) 5≠10 *

   Solving this system of equations resulted in a contradiction, since 5 can never be equal to 10. Therefore, the lines are parallel and do not have a point of intersection.