Exercise 26 Page 229

In this system of equations, at least one of the variables has a coefficient of 1. 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 first equation.
   4y=x-8 & (I) - 14x+y=- 1 & (II)

   DivEqn

   (I): .LHS /4.=.RHS /4.

   4y4= x-84 - 14x+y=- 1

   CancelCommonFac

   (I): Cancel out common factors

   4y4= x-84 - 14x+y=- 1

   SimpQuot

   (I): Simplify quotient

   y= x-84 - 14x+y=- 1

   WriteDiffFrac

   (I): Write as a difference of fractions

   y= x4- 84 - 14x+y=- 1

   CalcQuot

   (I): Calculate quotient

   y= x4-2 - 14x+y=- 1

   MoveNumRight

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

   y= 14x-2 - 14x+y=- 1
   Now that the y-variable is isolated in Equation (I), we can substitute its equivalent expression in Equation (II).
   y= 14x-2 & (I) - 14x+y=- 1 & (II)

   Substitute

   (II): y= 1/4x-2

   y= 14x-2 - 14x+( 14x-2)=- 1

   RemovePar

   (II): Remove parentheses

   y= 14x-2 - 14x+ 14x-2=- 1

   AddTerms

   (II): Add terms

   y= 14x-2 - 2≠ - 1 *
   Since we obtained a false statement, the system has no solutions.