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{[(1)/(4)x-(1)/(3)y=(3)/(2)],[(1)/(4)x+(3)/(4)y=(5)/(2)]:}

$\left\{\begin{array}{l}\frac{1}{4} x-\frac{1}{3} y=\frac{3}{2} \\ \frac{1}{4} x+\frac{3}{4} y=\frac{5}{2}\end{array}\right.$

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Multiplication with rational exponents

Full solution

Q.

\left\{\begin{array}{l}\frac{1}{4} x-\frac{1}{3} y=\frac{3}{2} \\ \frac{1}{4} x+\frac{3}{4} y=\frac{5}{2}\end{array}\right.

Write Equations: Step $1$ : Write down the system of equations. $\newline$ $(\frac{1}{4})x - (\frac{1}{3})y = \frac{3}{2}$ , $(\frac{1}{4})x + (\frac{3}{4})y = \frac{5}{2}$
Clear Fractions: Step $2$ : Multiply the first equation by $12$ to clear the fractions. $\newline$ $12 \times \left[\left(\frac{1}{4}\right)x - \left(\frac{1}{3}\right)y = \frac{3}{2}\right]$ $\newline$ $3x - 4y = 18$
Solve for x: Step $3$ : Multiply the second equation by $4$ to clear the fractions. $\newline$ $4 \times \left[\left(\frac{1}{4}\right)x + \left(\frac{3}{4}\right)y = \frac{5}{2}\right]$ $\newline$ $x + 3y = 10$
Solve for y: Step $4$ : Solve for x from the second equation. $\newline$ $x = 10 - 3y$
Substitute x: Step $5$ : Substitute x in the first equation. $\newline$ $3(10 - 3y) - 4y = 18$ $\newline$ $30 - 9y - 4y = 18$ $\newline$ $30 - 13y = 18$
Solve for y: Step $6$ : Solve for y. $\newline$ $-13y = 18 - 30$ $\newline$ $-13y = -12$ $\newline$ $y = -12 / -13$ $\newline$ $y = \frac{12}{13}$
Substitute y: Step $7$ : Substitute $y$ back into the equation for $x$ . $x = 10 - 3\left(\frac{12}{13}\right)$ $x = 10 - \frac{36}{13}$ $x = \frac{130}{13} - \frac{36}{13}$ $x = \frac{94}{13}$

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