Core Connections Geometry, 2013
CC
Core Connections Geometry, 2013 View details
1. Section 11.1
Continue to next subchapter

Exercise 79 Page 698

Practice makes perfect
a Even though both equations have a variable with a coefficient of 1, the Substitution Method may not be the easiest. Instead, we will use the Elimination Method.

To solve a system using this method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other equation. This means that either the x-terms or the y-terms must cancel each other out. y+3 x=14 & (I) y-3 x=6 & (II)We can see that the x-terms will eliminate each other if we add Equation (II) to Equation (I).
y+3x=14 y-3x=6
y+3x+( y-3x)=14+ 6 y-3x=6
â–Ľ
(II):Solve for x
2y=20 y-3x=6
y=10 y-3x=6
Now we can solve for x by substituting the value of y into either equation and simplifying.
y=10 y-3x=6
y=10 10-3x=6
â–Ľ
(II): Solve for x
y=10 10=6+3x
y=10 4=3x
y=10 43=x
y=10 x= 43
The solution, or point of intersection, of the system of equations is ( 43, 10).
b

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 using this 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=6-3x & (I) 2x+y=7 & (II)
    y=x+5 2x+ x+5=7
    y=x+5 3x+5=7
    y=x+5 3x=2
    y=x+5 x= 23
    Great! Now, to find the value of y, we need to substitute x=5 into either one of the equations in the given system. Let's use the first equation.
    y=x+5 x= 23
    y= 23+5 x= 23
    y= 23+ 153 x= 23
    y= 173 x= 23
    The solution, or point of intersection, to this system of equations is the point ( 23, 173).