Exercise 13 Page 265

In this system of equations, at least one of the variables terms is already isolated. Therefore, we will approach its solution with the Substitution Method. 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 second equation.
   -4x+4y=32 & (I) 3x+24=3y & (II)

   DivEqn

   (II): .LHS /3.=.RHS /3.

   -4x+4y=32 & (I) x+8=y & (II)

   RearrangeEqn

   (II): Rearrange equation

   -4x+4y=32 & (I) y=x+8 & (II)
   Now that we have isolated y, we can solve the system by substitution.
   -4x+4y=32 & (I) y=x+8 & (II)

   Substitute

   (I): y= x+8

   -4x+4( x+8)=32 & (I) y=x+8 & (II)

   Distr

   (I): Distribute 4

   -4x+4x+32=32 & (I) y=x+8 & (II)

   AddTerms

   (I): Add terms

   32=32 & (I) y=x+8 & (II)
   Solving this system of equations resulted in an identity; 32 is always equal to itself. Therefore, the lines are the same and have infinitely many intersection points.