Exercise 1 Page 211

We will solve the given system of equation using substitution. In this method there are three steps to follow.

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+4 & (I) x+y=- 12 & (II)

   Substitute

   (II):y= x+4

   y=x+4 x+ x+4=- 12

   AddTerms

   (II):Add terms

   y=x+4 2x+4=- 12

   SubEqn

   (II):LHS-4=RHS-4

   y=x+4 2x=- 16

   DivEqn

   (II): .LHS /2.=.RHS /2.

   y=x+4 x=- 8

   Great! Now, to find the value of y, we need to substitute x=- 8 into Equation (I).

   y=x+4 x=- 8

   Substitute

   (I):x= -8

   y= -8+4 x=-8

   AddTerms

   (I): Add terms

   y=-4 x=-8

   The solution, or point of intersection, to this system of equations is the point (-8,-4).