Question

Solve the system:
x^2 +y^2 =289
2x^2 -y^2 =143
The solutions will have the form (+-a, +- sqrt b) What is b?

A. 17
B. 145
C. 3
D. -37
E. none of these

Answer 1

Do some substitution.

Answer 2

What should i substitute? can you walk me through this?

Answer 3

Just solve this as you would any other system of equations, except this time you will have 2 answers.

Answer 4

\[x^2+y^2 = 289\]\[2x^2-y^2=143\]You can see by the process of elimination that the y's can cancel eachother out. You'd be left with:\[3x^2=289+143\]\[3x^2=432\]\[x^2=144\]\[x= \pm 12\] Now we substitute the x values into the first equation: \(x^2+y^2=289\) to find our y value. Here, it does not matter whether you choose to use \(-12\) or \(+12\) because You will end up with a positive value when you substitute it in.

Answer 5

So...\[(12)^2 +y^2 = 289\]\[144 +y^2 = 289\]\[y^2 = 289-144 = 145\]\[y = \pm \sqrt{145}\]

Answer 6

So now it's evident what your solution can be.

Answer 7

Okay thanks so much! that makes a lot of sense.