Solve the following pairs of equations:4x + f | vedclass

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(A) Given pair of linear equations are:$4x + \frac{6}{y} = 15$ ..... $(i)$$6x - \frac{8}{y} = 14, y 
eq 0$ ..... $(ii)$Let $u = \frac{1}{y}$,then the

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$3, 2$

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(A) Given pair of linear equations are:$4x + \frac{6}{y} = 15$ ..... $(i)$$6x - \frac{8}{y} = 14, y 
eq 0$ ..... $(ii)$Let $u = \frac{1}{y}$,then the equations become:$4x + 6u = 15$ ..... $(iii)$$6x - 8u = 14$ ..... $(iv)$To eliminate $u$,multiply equation $(iii)$ by $4$ and equation $(iv)$ by $3$:$16x + 24u = 60$ ..... $(v)$$18x - 24u = 42$ ..... $(vi)$Adding equation $(v)$ and $(vi)$:$(16x + 18x) + (24u - 24u) = 60 + 42$$34x = 102$$x = \frac{102}{34} = 3$Substitute $x = 3$ in equation $(iii)$:$4(3) + 6u = 15$$12 + 6u = 15$$6u = 15 - 12 = 3$$u = \frac{3}{6} = \frac{1}{2}$Since $u = \frac{1}{y}$,we have $\frac{1}{y} = \frac{1}{2}$,which implies $y = 2$.Thus,the solution is $x = 3$ and $y = 2$.

Solve the following pairs of equations:
$4x + \frac{6}{y} = 15$
$6x - \frac{8}{y} = 14, y \neq 0$

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Solve the following pairs of equations:$4x + \frac{6}{y} = 15$$6x - \frac{8}{y} = 14, y \neq 0$