Question

Examine this system of equations. Which number can be multiplied by each equation so that when the two equations are

Answer 1

Answer:

15 times the first equation and -12 times the second equation

Step-by-step explanation:

we have

$\frac{3}{4}x+\frac{2}{3}y=6$ ------> first equation

$\frac{5}{8}x+\frac{5}{6}y=12$ ------> second equation

Multiply the first equation by 15 both sides

$(15)\frac{3}{4}x+(15)\frac{2}{3}y=6(15)$

$\frac{45}{4}x+10y=90$ -----> new first equation

Multiply the second equation by -12 both sides

$(-12)\frac{5}{8}x+(-12)\frac{5}{6}y=12(-12)$

$-\frac{60}{8}x-10y=-144$ -----> new second equation

Adds the two new equations

$\frac{45}{4}x+10y=90\\-\frac{60}{8}x-10y=-144\\-----------\\\frac{45}{4}x-\frac{60}{8}x=90-144$

The y-term was eliminated