Question

Can someone help me solve these systems using elimination? I have a hard time with these.

Answer 1

Question 1:

For this case we must solve the following system of equations by the method of elimination:

$8x-6y=-20\\-16x+17y=30$

We multiply the first equation by 2:

$16x-12y = -40$

We have the equivalent system:

$16x-12y = -40\\-16x + 17y = 30$

We add the equations:

$5y = -10$

We divide between 5 on both sides of the equation:

$y = - \frac {10} {5}\\y = -2$

We find the value of the variable "x":

$8x-6 (-2) = - 20\\8x + 12 = -20\\8x = -20-12\\8x = -32$

We divide by 8 on both sides of the equation:

$x = - \frac {32} {8}\\x = -4$

Thus, the solution of the system is:

$(x, y): (- 4, -2)$

Answer:

$(x, y): (- 4, -2)$

Question 2:

For this case we must solve the following system of equations by the method of elimination:

$-4y-11x = 36\\20 = -10x-10y$

We multiply the first equation by 10:

$-40y-110x = 360$

We multiply the second equation by -4:

$40y + 40x = -80$

We have the equivalent system:

$-40y-110x = 360\\40y + 40x = -80$

We add the equations:$-70x = 280$

We divide by -70 on both sides of the equation:

$x = \frac {280} {- 70}\\x = -4$

We find the value of the variable "y":

$-4y-11 (-4) = 36\\-4y + 44 = 36\\-4y = 36-44\\-4y = -8$

We divide between -4 on both sides of the equation:

$y = - \frac {8} {- 4}\\y = 2$

Thus, the solution of the system is:

$(x, y): (- 4,2)$

Answer:

$(x, y): (- 4,2)$