Question

what is one of the solutions to the following system of equations y^2+x^2=53 y-x=5

Answer 1

rewrite the question please

Answer 2

y^2 +x^2=53 and y-x=5

Answer 3

what is one of the solutions to the following system of equations

Answer 4

@mayafrancis

Answer 5

\[y^2+x^2=53\]\[y-x=5\]

To solve this, solve the second equation for either \(x\) or \(y\). Then substitute that expression in place of the variable you solved for in the first equation. Now you should have an equation in terms of just one variable which you can solve. When you've done that, plug the value(s) found back into the second equation to find the value of the other variable.

Answer 6

So like y=5+x?

Answer 7

yes. now put that in the first first eqn

Answer 8

where you have y replace it with (5 + x)

Answer 9

do you need more help? did you get 2, -7

Answer 10

\(y=5+x\) would be a fine start. then in the first equation, wherever you see \(y\), replace it with \((5+x)\) and solve.

Answer 11

when you've solved that equation for \(x\) (there will be two solutions, because it has \(x^2\) as the highest power of \(x\)), take your solutions and use them in \(y = 5+x\) to find the corresponding values of \(y\).

Answer 12

y=5+x +x^2 = 53?

Answer 13

y^2 + x^2 =53
(5+x)^2 +x^2=53

Answer 14

Hmm idk how to solve that

Answer 15

yes you do
i'll work with you

Answer 16

First step is to expand it out:

\[(x+5)^2 + x^2 = 53\]\[(x+5)(x+5) + x^2 = 53\]
What do you get when you do the multiplication?

Answer 17

ok let me see

Answer 18

x2+10x+25

Answer 19

now add the x^2 and set = 53

Answer 20

what do you have

Answer 21

wait add x2 to wat

Answer 22

look at the original Y^2 + x^2 = 53
then you substituted for y so you get
(x+5)^2 + x^2 = 53
you expanded (x+5)^2 to get X^2 + 10x + 25 so now you have
x^2 +10x +25 + x^2 = 53

Answer 23

now group like terms and set = 0

Answer 24

2x^2+10x+25=53

Answer 25

and then minus 25?

Answer 26

can i plz have the answer lol i gotta get this done

Answer 27

\[2x^2+10x + 25 = 53\]Subtract 53 from both sides to get in quadratic formula arrangement.\[2x^2+10x+25-53=0\]\[2x^2+10x-28 = 0\]\[a=2,b=10,c=-28\]\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

Answer 28

-124?

Answer 29

helloooo

Answer 30

2X^2 + 10X-28 =0
DIVIDE BY 2
X^2 + 5X-14 = 0
2(X+7)(X-2) =0
X=-7 X=2
you have x values find the corresponding y
remember y= x +5
when x = -7, y = -7+5 = -2
when x = 2, y = 2 + 5 = 7

Answer 31

one solution (2 ,7)

Answer 32

to check in eqn 1
left side x^2 + y^2 = 2^2 + 7^2 = 4 + 49 = 53
right side 53
53 = 53 x=2, y = 7 is a solution

Answer 33

Thank you so much ::)

Answer 34

welcome

Answer 35

\[a=2, b= 10, c = -28\]\[x=\frac{-b\pm\sqrt{b^2 - 4 a c}}{2a} = \frac{-10\pm\sqrt{10^2-4*2*(-28)}}{2*2} \]\[=\frac{1}{4}*(-10\pm\sqrt{100+224})\]\[=\frac{1}4*(-10\pm\sqrt{324}) = \frac{1}{4}(-10\pm18)\]\[x=2 \rightarrow y = x+5 = 2+5 = 7 \rightarrow (2,7)\text { is a solution}\]\[x=-7\rightarrow y = -7+5=-2\rightarrow(-7, -2)\text{ is a solution}\]

For this equation, factoring is probably faster for most. For a less friendly one, or one which couldn't be factored, the quadratic formula becomes more appealing.