Question

x^2-4^2=9
y-2x=0 solve by substitution method..can someone show me the steps

Answer 1

\(y-2x=0\)
Isolate y;
\(y=2x\)

Answer 2

Is it x^2-4\(\huge y\)^2=9?

Answer 3

If so, substitute y into the first equation you got

\(\begin{align} \\
x^2-4y^2&=9 \\
x^2-4(2x)^2&=9 \\
x^2-4(4x^2)&=9 \\
x^2-16x^2 &=9 \\
x^2-16x^2 &=9 \\
-15x^2&=9 \\
x^2&=0.6 \\
x&=\pm \sqrt{0.6}



\end{align}\)

Answer 4

the book says that the answer is none tho

Answer 5

but i don't how they got that answer

Answer 6

Are you sure the question is \(x^2-4^2=9 \\
y-2x=0\)
?

Answer 7

zepp

you go from

-15x^2=9

to

x^2=0.6


but you lost the sign somehwere as it should be x^2=-0.6

Answer 8

yes I am sure

Answer 9

somewhere*

Answer 10

so that explains why there are no real answers

Answer 11

Ohh, my minus :( Where are you??

\(\pm \sqrt{-0.6}\) then, therefore no real solution.

Answer 12

but asker states that's \(x^2−4^2=9\) and not \(x^2-4y^2=9\)

Answer 13

Well if the first equation is x^2−4^2=9, then there are 2 ordered pair real solutions.

But the book states that there are no real solutions.

So my guess is that it should be x^2−4y^2=9

Answer 14

the problem is x^2-4^2=9
y-2x=0 i looked again

Answer 15

then there's a typo in the problem or there's a mistake in the answer given

Answer 16

\(x^2-4^2=9\\x^2-16=9\\x^2 =9+16\\x^2=25\\x=\pm\sqrt{25}\)

\(x=5;-5\)


\(y-2x=0\\y-2(5)=0\\y-10=0\\y=10\)

Ordered pair 1: \((5,10)\)

\(y-2x=0\\y-2(-5)=0\\y+10=0\\y=-10\)

Ordered pair 1: \((-5,-10)\)

Answer 17

There must be a typo in the book :)

Answer 18

ok so, the last thing u sent me is how u solve the problem by the substitution method?

Answer 19

Yes, I found x, and substituted x into the second equation to find y.

Answer 20

ok it, thanks!

Answer 21

You are welcome :)