Big Ideas Math Algebra 1, 2015
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Big Ideas Math Algebra 1, 2015 View details
Chapter Review
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Exercise 12 Page 283

How can we manipulate the equations to make a variable term cancel out the corresponding variable term in the other equation?

No solution.

Practice makes perfect
Since neither equation has a variable with a coefficient of 1 to use the Substitution Method; we will use the Elimination Method. 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 equation. This means that either the x-terms or the y-terms must cancel each other out. 3 x-6 y=-9 & (I) -5 x+10 y=10 & (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 5 and multiply Equation (II) by 3, the x-terms and y-terms will have opposite coefficients. 5(3 x-6 y)=5(-9) & (I) 3(-5 x+10 y)=3(10) & (II) ⇓ 15x- 30y=-45 & (I) -15x+ 30y=30 & (II) We can see that the x-terms and y-terms will eliminate each other if we add Equation (I) to Equation (II).
15x-30y=- 45 -15x+30y=30
15x-30y=- 45 -15x+30y+( 15x-30y)=30+( -45)
15x-30y=- 45 0=-15 *
Solving this system of equations resulted in a contradiction; 0 can never be equal to -15. The lines are parallel and do not have a point of intersection. Therefore, the system of linear equations has no solution.