Big Ideas Math: Modeling Real Life, Grade 8
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Big Ideas Math: Modeling Real Life, Grade 8 View details
3. Solving Systems of Linear Equations by Elimination
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Exercise 29 Page 217

Are there any variables already isolated?

Solution: (-14/5,-8/5)
Method: See solution.

Practice makes perfect
Since the variable x in the first equation has a coefficient equal to 1, we can isolate it using the Properties of Equality.
2=x-3y & (I) -2x+y=4 & (II)
2+3y=x-3y+3y -2x+y=4
2+3y=x -2x+y=4
x=3y+2 -2x+y=4
Now that x is isolated, the easiest way to solve the system of equations is to use the Substitution Method. Let's substitute x=3y+2 into the second equation.
x=3y+2 -2x+y=4
x=3y+2 -2( 3y+2)+y=4
â–Ľ
(II):Solve for y
x=3y+2 (-2)3y+(-2)2+y=4
x=3y+2 -6y+(-4)+y=4
x=3y+2 -6y-4+y=4
x=3y+2 -6y-4+y+4=4+4
x=3y+2 -5y=8
x=3y+2 -5y/-5=8/-5
x=3y+2 -5/-5 * y=8/-5
x=3y+2 5/5 * y=8/-5
x=3y+2 1 * y=8/-5
x=3y+2 y=8/-5
x=3y+2 y=-8/5
Now we can solve for x by substituting the value of y into either equation and simplifying.
x=3y+2 y=-8/5
x=3( -8/5)+2 y=-8/5
â–Ľ
(I):Solve for y
x=-3 * 8/5+2 y=-8/5
x=-3 * 8/5+2 y=-8/5
x=-3 * 8/5+5 * 2/5 y=-8/5
x=-24/5+10/5 y=-8/5
x=-24+10/5 y=-8/5
x=-14/5 y=-8/5
x=-14/5 y=-8/5
The solution, or point of intersection, of the system of equations is (- 145,- 85).