system of equation y=x^2+3 y=x+5

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system of equation y=x^2+3 y=x+5

Mathematics
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if anyone could show me the steps?
We generally use one to solve the other. Start with simplest. y = x + 5 y(-5) = x+ 5 (-5) y- 5 = x therefore substituting in eq .1 we get: y = (y-5)^2 +3 y = (y^2 -10y +25) + 3 y = y^2 -10y + 22 subtract y from both sides to get: 0 = y^2 -9y + 22 solve to get: |dw:1439459437832:dw|
|dw:1439459686316:dw|

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Other answers:

so @BPDlkeme234 so the answer is 9 +or- 72over 22
y = (y-5)^2 +3 y = (y^2 -10y +25) + 3 this should actually come to y^2-10y-y+25+3=0 y^2-11y+28=0 can you finish solving for y y^2-11y+28 is factorable
@myininaya how would I factor y^2
why would you want to factor y^2?
y^2 is already in factored form... I was wanting you to factor y^2-11y+28
what two numbers multiply to be 28 and add up to be -11?
y7 and y4 @myininaya
-7(-4)=28 -7-4=-11 so y^2-11y+28=(y-7)(y-4)
so you can solve (y-7)(y-4)=0 by solving both y-7=0 or y-4=0
once you find the y's then plug each y into: \[x=y-5\]
I still do not understand why im sorry im vary lost @myininaya
understand what exactly?
factoring?
if you don't understand how to factor you can use the quadratic formula since it is a quadratic equation
\[ax^2+bx+c=0 \\ \implies x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\]
\[y^2-11y+28=0 \\ \implies y=\frac{-(-11) \pm \sqrt{(-11)^2-4(1)(28)}}{2(1)}\] do you understand how to simplify this?
I don't know how to find the answer I thought he answer was y7 and x4
before we can find the ordered pairs we have to find the y values do you know how to solve the equation y^2-11y+28=0?
when you say things like y7 does that mean y=7?
if so when you said "y7 and y4 " which actually in your case means y=7 or y=4 then that is right for the y's... now you had x=y-5 now if y=7 then x=7-5=? and if y=4 then x=4-5=? so the ordered pairs are (7-5,7) (4-5,4)
just need to simplify the x-coordinates for each point

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