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Mathematics
carmelle:

How do I solve these two system of equations algebraically? y = x^2 + 3x + 5 y = x + 13

katelyn865:

Step 1: Add the two equations. Step 2: Solve for x. Step 3: To find the y-value, substitute in 3 for x in one of the equations. Step 4: Solve for y.

carmelle:

._. huh?

katelyn865:

thats how

carmelle:

im confused-

katelyn865:

sorry i realy dont no these good

carmelle:

bruh

katelyn865:

|dw:1619126027525:dw|

katelyn865:

|dw:1619126039938:dw|

carmelle:

stop spamming

katelyn865:

sorrry

snowflake0531:

We can do this the same we do linear equations \[y=x^2+3x+5\] \[y=x+13\] We have \[x^2+3x+5=x+13\] can you get them to one side of the equation and factor it?

carmelle:

I can try ;-;

katelyn865:

smart

katelyn865:

you got this

carmelle:

actually no, I cant

snowflake0531:

... just get them to one side of the equation make it __________=0

carmelle:

ohhh

carmelle:

x^2 + 2x - 8 = 0

snowflake0531:

Yes so I know that you're terrible at factoring, and I don't feel like being in the mood to spend the time to talk we have (x-4)(x+2) which makes x =4 and x=-2

carmelle:

thank youuuu

snowflake0531:

that isn't hte answer... now find y

carmelle:

how do I find y :I

snowflake0531:

You have (4,_) and (-2,_) so plug these x values into y=x+13

carmelle:

ohhh y = 4 + 13 y = 17 ..?

snowflake0531:

Yes so (4,13) and now do it for -2

snowflake0531:

Wait wait wait shoot

snowflake0531:

(-4,_) and (2,_)

snowflake0531:

So plug those 2 into y=x+13

carmelle:

ah ok, y = 15

snowflake0531:

Yes that's for 2, how about -4

carmelle:

and y = 9

snowflake0531:

Yes so there are your answers (-4,9) and (2,15)

carmelle:

yay tysm

snowflake0531:

yw~

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