Question

practice using linear combinations to solve systems of equations.which constants can be multiplied by the equations so one variable will be eliminated when the systems are added together?5x + 13y = 23212x + 7y = 218the first equation can be multiplied by -13 and the second equation by 7 to eliminate ythe first equation can be multiplied by 7 and the second equation by 13 to eliminate ythe first equation can be multiplied by -12 and the second equation by 5 to eliminate xthe first equation can be multiplied by 5 and the second equation by 12 to eliminate x

Explanation:

Step1: Test eliminating y (Option1)

Multiply eq1 by -13: $-13(5x+13y)=-13(232) \implies -65x-169y=-3016$
Multiply eq2 by 7: $7(12x+7y)=7(218) \implies 84x+49y=1526$
Y-terms: $-169y+49y=-120y$ (not eliminated)

Step2: Test eliminating y (Option2)

Multiply eq1 by 7: $7(5x+13y)=7(232) \implies 35x+91y=1624$
Multiply eq2 by 13: $13(12x+7y)=13(218) \implies 156x+91y=2834$
Y-terms: $91y+91y=182y$ (not eliminated)

Step3: Test eliminating x (Option3)

Multiply eq1 by -12: $-12(5x+13y)=-12(232) \implies -60x-156y=-2784$
Multiply eq2 by 5: $5(12x+7y)=5(218) \implies 60x+35y=1090$
X-terms: $-60x+60x=0$ (eliminated)

Step4: Test eliminating x (Option4)

Multiply eq1 by 5: $5(5x+13y)=5(232) \implies 25x+65y=1160$
Multiply eq2 by 12: $12(12x+7y)=12(218) \implies 144x+84y=2616$
X-terms: $25x+144x=169x$ (not eliminated)

Answer:

C. The first equation can be multiplied by -12 and the second equation by 5 to eliminate x