Problem

E. $\quad x \geq 0$
Question 7
The solution of the two simultaneous equations $2 a x+2 b y=3$ and $3 a x-2 b y=7$ for $x$ and $y$ is:
A. $x=2 a, y=\frac{3-4 a^{2}}{2 b}$
B. $x=2, y=\frac{3-4 a}{2 b}$
C. $x=0, y=0$
D. $x=3 a, y=7 b$
E. $x=\frac{2}{a}, y=-\frac{1}{2 b}$
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Answer

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Answer

The solution is \(\boxed{x = \frac{3 + b^2}{2a}, y = -\frac{1}{2}b}\)

Steps

Step 1 :Multiply the first equation by 3 and the second equation by 2 to make the coefficients of y equal:

Step 2 :6ax + 6by = 9,

Step 3 :6ax - 4by = 14.

Step 4 :Subtract the second equation from the first equation:

Step 5 :10by = -5 \Rightarrow y = -\frac{1}{2}b.

Step 6 :Substitute y back into the first equation:

Step 7 :2ax + 2b\left(-\frac{1}{2}b\right) = 3 \Rightarrow 2ax - b^2 = 3.

Step 8 :Divide both sides by 2a:

Step 9 :x = \frac{3 + b^2}{2a}.

Step 10 :The solution is \(\boxed{x = \frac{3 + b^2}{2a}, y = -\frac{1}{2}b}\)

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