Question

Is the answer a, b, c, or d.

Answer 1

The correct answer is d.

We have the following system of linear equations:

(I) $4x+7y=-14$
(II) $8x+5y=8$

Let's use the elimination method, then let's multiply the equation (1) $\times(-2)$ and subtracting (I) and (II):

(I) $-2(4x+7y)=-2(-14)$
∴ $-8x-14y=28$

(I) $-8x-14y=28$
(II) $8x+5y=8$
____________________
(III)  $-9y=36$

∴  $y=-4$

We can find the value of x by substituting y either in (I) or (II). Thus, from (I):

$4x+7(-4)=-14$
∴ $4x-28=-14$
∴ $4x=14$
∴ $x=\frac{14}{4}$
∴ $\boxed{x=\frac{7}{2}}$

Let's substitute the values of x and y into (I) and (2)

(I) $4(\frac{7}{2})+7(-4)=14-28=-14$
(II) $8(\frac{7}{2})+5(-4)=28-20=8$

Finally the answer is:

$\boxed{x=\frac{7}{2} \rightarrow x= 3 \frac{1}{2}}$