Exercise 3 Page 293

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. For this exercise, y is already isolated in one equation, so we can skip straight to solving!
   y=x+5 & (I) y=-3x+1 & (II)

   Substitute

   (II): y= x+5

   y=x+5 x+5=-3x+1

   ▼

   Solve for x

   AddEqn

   (II): LHS+3x=RHS+3x

   y=x+5 x+5+3x=-3x+1+3x

   AddTerms

   (II): Add terms

   y=x+5 4x+5=1

   SubEqn

   (II): LHS-5=RHS-5

   y=x+5 4x+5-5=1-5

   SubTerms

   (II): Subtract terms

   y=x+5 4x=-4

   DivEqn

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

   y=x+5 4x/4=-4/4

   MovePartNumRight

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

   y=x+5 4/4 * x=-4/4

   MoveNegNumToFrac

   (II): Put minus sign in front of fraction

   y=x+5 4/4 * x=-4/4

   QuotOne

   (II): a/a=1

   y=x+5 1 * x=-1

   IdPropMult

   (II): Identity Property of Multiplication

   y=x+5 x=-1
   Great! Now, to find the value of y, we need to substitute x=-1 into either one of the equations in the given system. Let's use the first equation.
   y=x+5 x=-1

   Substitute

   (I):x= -1

   y= -1+5 x=-1

   AddTerms

   (I): Add terms

   y=4 x=-1
   The solution, or point of intersection, to this system of equations is the point (-1,4).