McGraw Hill Glencoe Algebra 1, 2017
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McGraw Hill Glencoe Algebra 1, 2017 View details
5. Applying Systems of Linear Equations
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Exercise 1 Page 371

We are asked to solve the following system of equations. 2x+3y=-11 & (I) -8x-5y=9 & (II) Since neither equation has a variable with a coefficient of 1, the Substitution Method may not be the easiest. Instead, we will use the Elimination Method. To do that, 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- or the y-terms must cancel each other out. 2 x+3 y=-11 & (I) -8 x-5 y=9 & (II) Currently, none of the terms in this system will cancel out. Therefore, we need to find a common multiple between two variable like terms in the system. If we multiply Equation (I) by 4, the x-terms will have opposite coefficients. 4(2 x+3 y)=4(-11) -8 x-5 y=9 ⇓ 8x+12 y=-44 -8x-5 y=9We can see that the x-terms will eliminate each other if we add Equation (I) to Equation (II).
8x+12y=-44 -8x-5y=9
8x+12y=-44 -8x-5y+( 8x+12y)=9+( -44)
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(II):Solve for y
8x+12y=-44 -8x-5y+8x+12y=9-44
8x+12y=-44 -8x+8x+12y-5y=9-44
8x+12y=-44 7y=-35
8x+12y=-44 y=-5
Now we can now solve for x by substituting the value of y into either equation and simplifying.
8x+12y=-44 y=-5
8x+12( -5)=-44 y=-5
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(I):Solve for x
8x-60=-44 y=-5
8x=16 y=-5
x=2 y=-5
The solution, or point of intersection, of the system of equations is (2, -5).