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44. \\(\\frac{1}{2}x + \\frac{3}{4}y = \\frac{7}{4}\\)\\(\\frac{1}{3}x …

Question

  1. \\(\frac{1}{2}x + \frac{3}{4}y = \frac{7}{4}\\)\\(\frac{1}{3}x - \frac{1}{6}y = \frac{1}{2}\\)

Explanation:

Step1: Eliminate fractions by multiplying equations

Multiply the first equation \(\frac{1}{2}x+\frac{3}{4}y = \frac{7}{4}\) by 4 to get \(2x + 3y=7\).
Multiply the second equation \(\frac{1}{3}x-\frac{1}{6}y=\frac{1}{2}\) by 6 to get \(2x - y = 3\).

Step2: Subtract the two new equations

Subtract \(2x - y = 3\) from \(2x + 3y = 7\):
\((2x + 3y)-(2x - y)=7 - 3\)
Simplify: \(4y = 4\), so \(y = 1\).

Step3: Substitute \(y = 1\) into an equation

Substitute \(y = 1\) into \(2x - y = 3\):
\(2x-1 = 3\)
Add 1 to both sides: \(2x = 4\), so \(x = 2\).

Answer:

\(x = 2\), \(y = 1\)