Solve the following pair of linear equations | vedclass

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(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$.

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$(2, -3)$

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(D) To solve the system of equations $x + 2y = -4$ and $3x + 4y = -6$ graphically:$1$. For the first equation $x + 2y = -4$,if $x = 0$,then $y = -2$. If $y = 0$,then $x = -4$. The line passes through $(0, -2)$ and $(-4, 0)$.$2$. For the second equation $3x + 4y = -6$,if $x = 0$,then $y = -1.5$. If $y = 0$,then $x = -2$. The line passes through $(0, -1.5)$ and $(-2, 0)$.$3$. Solving the equations algebraically: Multiply the first equation by $2$: $2x + 4y = -8$. Subtract this from the second equation: $(3x + 4y) - (2x + 4y) = -6 - (-8)$,which gives $x = 2$.$4$. Substitute $x = 2$ into $x + 2y = -4$: $2 + 2y = -4 \implies 2y = -6 \implies y = -3$.$5$. The intersection point is $(2, -3)$.

Solve the following pair of linear equations in two variables using a graph: $x + 2y = -4$ and $3x + 4y = -6$.

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