Question

Solve by substitution
xy-4=0
x^2+4y^2=20

Answer 1

Solve one equation for one variable. Then, replace a variable in the second equation with the variable that you solved in the first equation.
For example, solving x+y=5 and xy=6, I could solve the first equation for a variable so that I get x=5-y and xy=6. Then, I replace the x in the second equation (since that's the variable that I solved for) so that we get one equation (5-y)y=6. Then I solve for y.

Answer 2

I know but I get stuck

Answer 3

on the variable in the denominator part

Answer 4

I solve the first equation for x = (4/y)

then I plug it into the second equation

(4/y)^2 + 4y^2 = 20

(16/y^2)+4y^2=20

Then I don't know what to do

Answer 5

At this point you can find the least common denominator of 16/y^2 and (4y^2)/1. And then add them together.

Answer 6

how?

Answer 7

can you do it first, then explain it please?

Answer 8

usually I only need to see work and a little explanation...

Answer 9

I will show you a different example so you can learn how to do it yourself. Let's look at \[\frac{ 3 }{ x }+4\]To simplify this, you need to add the fractions, but fractions can only be added together when they have like denominators. So, since x is the LCD (x is the lowest number that the denominator of each fraction can go into), you multiply the numerator and denominator of each fraction by what gets the denominator to the LCD. So in my example, you would multiply the first fraction by 1/1 and the second fraction by x/x to get \[\frac{ 3 }{ x }+\frac{ 4x }{ x }\] Now you can add these together to get \[\frac{ 3+4x }{ x }\]

Answer 10

you have got to be kidding me. I asked for this QUESTION SPECIFICALLY. I did NOT ask for THAT question. I am STUCK ON THIS QUESTION! Stop trying to be a teacher and just show me how to do this. You're wasting my time and yours.

Answer 11

Also, that is a horrible method of teaching. People need to be shown how to do that specific problem, then explain what you did!

Answer 12

I am tired of people on this site constantly doing that.

Answer 13

Your LCD is 4y^2, so multiply each numerator and denominator of 4y^2 by that.

Answer 14

is it 16/y^2 + 4y^4 / y^2 = 20y^2?

Answer 15

You won't do it to the 20 too, only to the fractions that you're adding. But yes the left side looks right.

Answer 16

but wouldn't I get (16+4y^4)/y^2?

Answer 17

Yes, but now you're closer to solving it as you can multiply each side of the equation by y^2. You can then set it equal to 0 and solve like you would a quadratic equation.

Answer 18

OOOOOOh I see! Thank you!

Answer 19

Happy to help. Oh and I only teach that way to inspire thought because many people only care about an answer and if I only walk them through it they might ignore the process.