Exercise 50 Page 384

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!
   y= 12x & (I) 2y+3x=28 & (II)

   Substitute

   (II):y= 1/2x

   y= 12x 2( 12x)+3x=28

   FracMultDenomToNumber

   (II):a/2* 2 = a

   y= 12x x+3x=28

   AddTerms

   (II):Add terms

   y= 12x 4x=28

   DivEqn

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

   y= 12x x=7

   Great! Now, to find the value of y, we need to substitute x for 7 into the first equation in the given system.

   y= 12x x=7

   Substitute

   (I):x= 7

   y= 12(7) x=7

   MoveRightFacToNumOne

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

   y= 72 x=7

   CalcQuot

   (I): Calculate quotient

   y=3.5 x=7

   The solution, or point of intersection, to this system of equations is the point (7,3.5).