Answer:
[tex]y=-6x+10[/tex]
Step-by-step explanation:
The point of intersection of
[tex]x+2y=9...eqn1[/tex]
and
[tex]4x-2y=-4...eqn2[/tex]
is the solution of the two equations.
We add equation (1) and equation(2) to get,
[tex]x+4x+2y-2y=9+-4[/tex]
[tex]\Rightarrow 5x=5[/tex]
[tex]\Rightarrow x=1[/tex]
We put [tex]x=1[/tex] into equation (1) to get,
[tex]1+2y=9[/tex]
[tex]\Rightarrow 2y=9-1[/tex]
[tex]\Rightarrow 2y=8[/tex]
[tex]\Rightarrow y=4[/tex]
Therefore the line passes through the point, [tex](1,4)[/tex].
The line also passes through the point of intersection of
[tex]3x-4y=14...eqn(3)[/tex]
and
[tex]3x+7y=-8...eqn(4)[/tex]
We subtract equation (3) from equation (4) to obtain,
[tex]3x-3x+7y--4y=-8-14[/tex]
[tex]\Rightarrow 11y=-22[/tex]
[tex]\Rightarrow y=-2[/tex]
We substitute this value into equation (4) to get,
[tex]3x+7(-2)=-8[/tex]
[tex]3x-14=-8[/tex]
[tex]3x=-8+14[/tex]
[tex]3x=6[/tex]
[tex]x=2[/tex]
The line also passes through
[tex](2,-2)[/tex]
The slope of the line is
[tex]slope=\frac{4--2}{1-2} =\frac{6}{-1}=-6[/tex]
The equation of the line is
[tex]y+2=-6(x-2)[/tex]
[tex]y+2=-6x+12[/tex]
[tex]y=-6x+10[/tex] is the required equation
