Core Connections Algebra 2, 2013
CC
Core Connections Algebra 2, 2013 View details
Chapter Closure

Exercise 117 Page 494

The Elimination Method can be used to solve a system of linear equations if either of the variable terms would cancel out the corresponding variable term in the other equation when added together.

(x,y) = (8,-3)

We want to solve the given system. x4 + y3 = 1 & (I) 2x - y3 = 17 & (II) Before finding the method to solve this, let's simplify the equations. We can get rid of the fractions by multiplying each equation by a common denominator of the fractions in it. Let's do it!
x4 + y3 = 1 & (I) 2x - y3 = 17 & (II)
3x + 4y = 12 2x - y3 = 17
3x + 4y = 12 6x - y = 51
To solve a system of linear equations using the Elimination Method, one of the variable terms needs to be eliminated when one equation is added to or subtracted from the other. This means that either the x- or the y-terms must cancel each other out. 3 x + 4 y = 12 & (I) 6 x - y = 51 & (II) Currently, none of the terms in this system will cancel out. Let's find a common multiple between two variable like terms in the system. If we multiply Equation (I) by -2, the x-terms will have opposite coefficients. -2(3 x + 4 y) = -2(12) 6 x - y = 51 ⇓ -6 x - 8 y = -24 6 x - y = 51 The x-terms will eliminate each other now if we add Equation (I) to Equation (II).
-6x - 8y = - 24 & (I) 6x - y = 51 & (II)
-6x - 8y = - 24 6x - y +( - 6x - 8y)= 51+( -24)
-6x - 8y = - 24 6x - y- 6x - 8y= 51-24
-6x - 8y = - 24 -9y= 27
-6x - 8y = - 24 y= - 3
We found that y=-3. Now we can solve for x by substituting this value of y into Equation (I) and simplifying. Let's do it!
-6x - 8y = - 24 & (I) y= - 3 & (II)
-6x - 8( -3) = - 24 y= - 3
(I):Solve for x
-6x + 24 = - 24 y= - 3
-6x = - 48 y= - 3
x = 8 y= - 3
The solution of the system of equations is (x,y) = (8,-3).