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(A) Given equations are:$(1)$ $x + 2y = \frac{3}{2}$$(2)$ $2x + y = \frac{3}{2}$To eliminate $y$,multiply equation $(1)$ by $1$ and equation $(2)$ by

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$\left(\frac{1}{2}, \frac{1}{2}\right)$

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(A) Given equations are:$(1)$ $x + 2y = \frac{3}{2}$$(2)$ $2x + y = \frac{3}{2}$To eliminate $y$,multiply equation $(1)$ by $1$ and equation $(2)$ by $2$:$x + 2y = \frac{3}{2}$$4x + 2y = 3$Subtracting the first equation from the second:$(4x - x) + (2y - 2y) = 3 - \frac{3}{2}$$3x = \frac{6 - 3}{2} = \frac{3}{2}$$x = \frac{3}{2 	imes 3} = \frac{1}{2}$Substitute $x = \frac{1}{2}$ into equation $(1)$:$\frac{1}{2} + 2y = \frac{3}{2}$$2y = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1$$y = \frac{1}{2}$Thus,the solution is $\left(\frac{1}{2}, \frac{1}{2}ight)$.

Solve the following pair of equations by the method of elimination:
$x + 2y = \frac{3}{2}$
$2x + y = \frac{3}{2}$

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Solve the following pair of equations by the method of elimination:$x + 2y = \frac{3}{2}$$2x + y = \frac{3}{2}$