Question 186832
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{{{4x + 5y = -2 }}}---->EQN 1
{{{4y - x = 11}}}-----> EQN 2
Based on Slope-Intercept Form:{{{system(y=mx+b)}}}
In  EQN 1: {{{5y=-4x-2}}}--->{{{cross(5)y/cross(5)=(-4x-2)/5}}}
{{{y=(-4/5)x-2/5}}}, EQN 1.1
Let fy=0
{{{0=(-4/5)x-2/5}}}---->{{{(4/5)x=-2/5}}}
{{{cross(4/5)x/cross(4/5)=(-2/5)/(4/5)=(-2/5)(5/4)}}}
{{{x=-2/4=-1/2}}}, X-Intercept
Let fx=0:
{{{y=(-4/5)(0)-2/5=-2/5}}}, Y-Intercept
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In EQN 2:
{{{4y=x+11}}}--->{{{cross(4)y/cross(4)=(x+11)/4}}}
{{{y=(1/4)x+11/4}}}, EQN 2.2
Let fy=0:----> {{{0=(1/4)x+11/4}}}
{{{(1/4)x=-11/4}}}---->{{{cross(1/4)x/cross(1/4)=(-11/4)/(1/4)}}}
{{{x=-11}}}, X-Intercept
Let fx=0:----> {{{y=(1/4)(0)+11/4=11/4}}}, Y-Intercept
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In EQN 2: By Substitution,
{{{4y-11=x}}}----->EQN 3, subst. in EQN 1
{{{4(4y-11)+5y=-2}}}
{{{16y-44+5y=-2}}}
{{{21y=-2+44=42}}}---->{{{cross(21)y/cross(21)=cross(42)2/cross(21)}}}
{{{highlight(y=2)}}}, Subst. in EQN 3:
{{{x=4*2-11=8-11}}}
{{{highlight(x=-3)}}}
POINT OF INTERSECTION (-3,2)
We'll see the graph:
{{{drawing(400,400,-12,8,-5,8,grid(1),graph(400,400,-12,8,-5,8,(-4/5)x-2/5,(1/4)x+11/4),blue(circle(-1/2,0,.2)),blue(circle(0,-2/5,.2)),blue(circle(-11,0,.2)),blue(circle(0,11/4,.2)),blue(circle(-3,2,.12)))}}}--->RED, EQN 1; GREEN, EQN 2
Thank you,
Jojo</pre>