Question

Solve the following using system of equations:
" x2 + y2 = 13
2x - y = 4
"

Answer 1

One of the things you can do here is rearrange the equations like this:
x2 +y2 -13 = 0
y = 2x + 4
After that what you can do is substitute:
x2 + (2x + 4)2 -13 = 0
x2 + (4x2 +16x + 16) -13 = 0
5x2 +16x + 3 = 0
x = (-16 + sqrt(16^2 - 4*5*3))/(2*5) and (-16 - sqrt(16^2 - 4*5*3))/(2*5)
x = (-16 + 14)/10 and (-16 - 14)/10
x = -0.2 and -3

y = 2x + 4
y = 2*(-0.2) + 4 = 3.6
y = 2*(-3) + 4 = -2
The answer will be
(-0.2,3.6) and (-3,-2)
Hope this helps you greatly

Answer 2

Answer: The value of (x,y)=(3,2) and

`(x,y)=((1)/(5),(-18)/(5))`

Explanation:

Since we have given that

`x^2+y^2=13\\\\2x-y=4`

From second equation i.e.

`2x-y=4`

`\implies 2x-4=y`

Put it in the first equation:

`x^2+y^2=13\\\\x^2+(2x-4)^2=13\\\\x^2+4x^2+16-16x=13\\\\5x^2-16x+16=13\\\\5x^2-16x=13-16\\\\5x^2-16x=-3\\\\5x^2-16x+3=0`

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)`

So, we will apply this to get the roots:

`\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x_(1,\:2)=(-b\pm √(b^2-4ac))/(2a)\\\\=(16+√(\left(-16\right)^2-4\cdot \:5\cdot \:3))/(2\cdot \:5)=3\\\\(16-√(\left(-16\right)^2-4\cdot \:5\cdot \:3))/(2\cdot \:5)=(1)/(5)`

Hence the value of x are

`(1)/(5),2`

Therefore, the value of y will be

`y=2x-4`

If we put
`x=(1)/(5)`

`y=2* (1)/(5)-4\\\\y=(2)/(5)-4\\\\y=(-18)/(4)`

If we put x=3

then

`y=2x-4\\\\y=2* 3-4\\\\y=6-4\\\\y=2`