The correct answer is d.
We have the following system of linear equations:
(I) [tex]4x+7y=-14[/tex]
(II) [tex]8x+5y=8[/tex]
Let's use the elimination method, then let's multiply the equation (1) [tex]\times(-2)[/tex] and subtracting (I) and (II):
(I) [tex]-2(4x+7y)=-2(-14)[/tex]
∴ [tex]-8x-14y=28[/tex]
(I) [tex]-8x-14y=28[/tex]
(II) [tex]8x+5y=8[/tex]
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(III) [tex]-9y=36[/tex]
∴ [tex]y=-4[/tex]
We can find the value of x by substituting y either in (I) or (II). Thus, from (I):
[tex]4x+7(-4)=-14[/tex]
∴ [tex]4x-28=-14[/tex]
∴ [tex]4x=14[/tex]
∴ [tex]x=\frac{14}{4}[/tex]
∴ [tex]\boxed{x=\frac{7}{2}}[/tex]
Let's substitute the values of x and y into (I) and (2)
(I) [tex]4(\frac{7}{2})+7(-4)=14-28=-14[/tex]
(II) [tex]8(\frac{7}{2})+5(-4)=28-20=8[/tex]
Finally the answer is:
[tex]\boxed{x=\frac{7}{2} \rightarrow x= 3 \frac{1}{2}}[/tex]
