Solve the following system:
{(3 x)/4 - y/3 = 2 | (equation 1)
x/2 + (2 y)/3 = 4 | (equation 2)
Subtract 2/3 × (equation 1) from equation 2:
{(3 x)/4 - y/3 = 2 | (equation 1)
0 x+(8 y)/9 = 8/3 | (equation 2)
Multiply equation 1 by 12:
{9 x - 4 y = 24 | (equation 1)
0 x+(8 y)/9 = 8/3 | (equation 2)
Multiply equation 2 by 9/8:
{9 x - 4 y = 24 | (equation 1)
0 x+y = 3 | (equation 2)
Add 4 × (equation 2) to equation 1:
{9 x+0 y = 36 | (equation 1)
0 x+y = 3 | (equation 2)
Divide equation 1 by 9:
{x+0 y = 4 | (equation 1)
0 x+y = 3 | (equation 2)
Collect results:
Answer: | x = 4 and y = 3
+14903 3/4x - 1/3y = 2 and 1/2x + 2/3 y = 4
\(\frac{3}{4}x-\frac{1}{3}y=2\\\frac{1}{2}x+\frac{2}{3}y=4\)
\(\frac{6}{4}x-\frac{2}{3}y=4\)
\(\frac{6}{4}x-4=\frac{2}{3}y\\4-\frac{1}{2}x=\frac{2}{3}y\)
\(\frac{6}{4}x-4=4-\frac{1}{2}x\\\frac{6}{4}x+\frac{2}{4}x=8\\2x=8\)
\(x=4\)
\(\frac{1}{3}y=\frac{3}{4}x-2\\y=\frac{9}{4}x-6=\frac{9}{4}\cdot 4-6\)
\(y=3\)
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