Find the equation of the line passing through | vedclass

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(C) Step $1$: Find the intersection point of $4x - 3y = 1$ and $5x - 2y = 3$.Multiplying the first equation by $2$ and the second by $3$: $8x - 6y = 2

Vedclass · Accepted Answer

$2x - 3y + 1 = 0$

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(C) Step $1$: Find the intersection point of $4x - 3y = 1$ and $5x - 2y = 3$.Multiplying the first equation by $2$ and the second by $3$: $8x - 6y = 2$ and $15x - 6y = 9$.Subtracting the equations: $7x = 7$,so $x = 1$.Substituting $x = 1$ into $4(1) - 3y = 1$,we get $3y = 3$,so $y = 1$.The intersection point is $(1, 1)$.Step $2$: The equation of a line parallel to $2x - 3y + 2 = 0$ is of the form $2x - 3y + k = 0$.Since the line passes through $(1, 1)$,we substitute the coordinates: $2(1) - 3(1) + k = 0$.$2 - 3 + k = 0$,which gives $k = 1$.Thus,the required equation is $2x - 3y + 1 = 0$.

Find the equation of the line passing through the intersection of the lines $4x - 3y - 1 = 0$ and $5x - 2y - 3 = 0$ and parallel to the line $2x - 3y + 2 = 0$.

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