y = x2 βˆ’ 6x + 82y + x = 4 The pair of points representing the solution set of this system of equations is

Accepted Solution

Answer:The two intersections are (3/2 , 5/4) and (4,0).I put two ways to do it. You can pick your favorite of these are try another route if you like. Step-by-step explanation:The system is:[tex]y=x^2-6x+8[/tex][tex]2y+x=4[/tex].I don't know how good at factoring you are but the top equation consists of polynomial expression that has a factor of (x-4). Β I see that if I solve 2y+x=4 for 2y I get 2y=-x+4 which is the opposite of (x-4) so -2y=x-4.So anyways, factoring x^2-6x+8=(x-4)(x-2) because -4+(-2)=-6 while -4(-2)=8.This is the system I'm looking at right now:[tex]y=(x-4)(x-2)[/tex][tex]-2y=x-4[/tex]I'm going to put -2y in for (x-4) in the first equation:[tex]y=-2y(x-2)[/tex]So one solution will occur when y is 0.Now assume y is not 0 and divide both sides by y:[tex]1=-2(x-2)[/tex]Distribute:[tex]1=-2x+4[/tex]Subtract 4 on both sides:[tex]-3=-2x[/tex]Divide both sides by -2:[tex]\frac{-3}{-2}=x[/tex][tex]\frac{3}{2}=x[/tex][tex]x=\frac{3}{2}[/tex]Now let's go back to one of the original equations:2y=-x+4Divide both sides by 2:[tex]y=\frac{-x+4}{2}[/tex]Plug in 3/2 for x:[tex]y=\frac{\frac{-3}{2}+4}{2}[/tex]Multiply top and bottom by 2:[tex]y=\frac{-3+8}{4}[/tex][tex]y=\frac{5}{4}[/tex]So one solution is at (3/2 , 5/4).The other solution happened at y=0:2y=-x+4Plug in 0 for y:2(0)=-x+40=-x+4Add x on both sides:x=4So the other point of intersection is (4,0).-------------------------------------------------------The two intersections are (3/2 , 5/4) and (4,0).Now if you don't like that way:[tex]y=x^2-6x+8[/tex][tex]2y+x=4[/tex]Replace y in bottom equation with (x^2-6x+8):[tex]2(x^2-6x+8)+x=4[/tex]Distribute:[tex]2x^2-12x+16+x=4[/tex]Subtract 4 on both sides:[tex]2x^2-12x+16+x-4=0[/tex]Combine like terms:[tex]2x^2-11x+12=0[/tex]Compare this to [tex]ax^2+bx+c=0[/tex][tex]a=2[/tex][tex]b=-11[/tex][tex]c=12[/tex]The quadratic formula is [tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]{Plug in our numbers:[tex]x=\frac{11 \pm \sqrt{(-11)^2-4(2)(12)}}{2(2)}[/tex][tex]x=\frac{11 \pm \sqrt{121-96}}{4}[/tex][tex]x=\frac{11 \pm \sqrt{25}}{4}[/tex][tex]x=\frac{11 \pm 5}{4}[/tex][tex]x=\frac{11+5}{4} \text{ or } \frac{11-5}{4}[/tex][tex]x=\frac{16}{4} \text{ or } \frac{6}{4}[/tex][tex]x=4 \text{ or } \frac{3}{2}[/tex]Using 2y+x=4 let's find the correspond y-coordinates.If x=4:2y+4=4Subtract 4 on both sides:2y=0Divide both sides by 2:y=0So we have (4,0) is a point of intersection.If x=3/22y+(3/2)=4Subtract (3/2) on both sides:2y=4-(3/2)2y=5/2Divide 2 on both sides:y=5/4The other intersection is (3/2 , 5/4).