Exercise 35 Page 28

Since the x-variable is isolated in the second equation, we will use the Substitution Method. To solve a system of linear equations this way, we have to follow 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. As we already know, x is isolated in one equation, so we can skip straight to solving!
   2x+2y=3 & (I) x=4y-1 & (II)

   Substitute

   (I): x= 4y-1

   2( 4y-1)+2y=3 & (I) x=4y-1 & (II)

   ▼

   (I):Solve for y

   Distr

   (I): Distribute 2

   8y-2+2y=3 & (I) x=4y-1 & (II)

   AddTerms

   (I): Add terms

   10y-2=3 & (I) x=4y-1 & (II)

   AddEqn

   (I):LHS+2=RHS+2

   10y=5 & (I) x=4y-1 & (II)

   DivEqn

   (I):.LHS /10.=.RHS /10.

   y= 510 & (I) x=4y-1 & (II)

   ReduceFrac

   (I):a/b=.a /5./.b /5.

   y= 12 & (I) x=4y-1 & (II)
   Great! Now, to find the value of x, we need to substitute y= 12 into either one of the equations in the given system. Let's use the second equation.
   y= 12 & (I) x=4y-1 & (II)

   Substitute

   (II):y= 1/2

   y= 12 & (I) x=4( 12)-1 & (II)

   ▼

   (II):Simplify right-hand side

   MoveLeftFacToNumOne

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

   y= 12 & (I) x= 42-1 & (II)

   CalcQuot

   (II): Calculate quotient

   y= 12 & (I) x=2-1 & (II)

   SubTerm

   (II): Subtract term

   y= 12 & (I) x=1 & (II)
   The solution, or point of intersection, to this system of equations is the point (1, 12).