Question

Y^2+X^2=36 y=-3x+5.

Answer 1

The two solutions in exact form are:

`((15+√(335))/(10),(5-3√(335))/(10))`

and

`((15-√(335))/(10),(5+3√(335))/(10))`.

If you prefer to look at approximations just put into your calculator:

`(3.3303,-4.9909)`

and

`(-0.3303,5.9909)`.

Explanation:

I guess you are asked to find the solution the given system.

I'm going to use substitution.

This means I'm going to plug the second equation into the first giving me:

`(-3x+5)^2+x^2=36` I replaced the 1st y with what the 2nd y equaled.

Before we continue solving this I'm going to expand the
`(-3x+5)^2` using the following:

`(a+b)^2=a^2+2ab+b^2`.

`(-3x+5)^2=(-3x)^2+2(-3x)(5)+(5)^2`

`(-3x+5)^2=9x^2-30x+25`

Let's go back to the equation we had:

`(-3x+5)^2+x^2=36`

After expansion of the squared binomial we have:

`9x^2-30x+25+x^2=36`

Combine like terms (doing the
`9x^2+x^2` part:

`10x^2-30x+25=36`

Subtract 36 on both sides:

`10x^2-30x+25-36=0`

Simplify the 25-36 part:

`10x^2-30x-11=0`

Compare this to
`ax^2+bx+c=0` which is standard form for a quadratic.

We should see the following:

`a=10`

`b=-30`

`c=-11`

The formula that solves this equation for the variable
`x` is:

`x=(-b \pm √(b^2-4ac))/(2a)`

Plugging in our values for
`a,b, \text{ and } c` give us:

`x=(30 \pm √((-30)^2-4(10)(-11)))/(2(10))`

Simplify the bottom; that is 2(10)=20:

`x=(30 \pm √((-30)^2-4(10)(-11)))/(20)`

Put the inside of square root into the calculator; that is put
`(-30)^2-4(10)(-11)` in the calculator:

`x=(30 \pm √(1340))/(20)`

Side notes before continuation:

Let's see if 1340 has a perfect square.

I know 1340 is divisible by 10 because it ends in 0.

1340=10(134)

134 is even so it is divisible by 2:

1340=10(2)(67)

1340=2(2)(5)(67)

1340=4(5)(67)

1340=4(335)

4 is a perfect square so we can simplify the square root part further:

`√(1340)=√(4)√(335)=2√(335)`.

Let's go back to the solution:

`x=(30 \pm √(1340))/(20)`

`x=(30 \pm 2 √(335))/(20)`

Now I see all three terms contain a common factor of 2 so I'm going to divide top and bottom by 2:

`x=((30)/(2) \pm (2 √(335))/(2))/((20)/(2))`

`x=(15 \pm √(335))/(10)`

So we have these two x values:

`x=(15+√(335))/(10) \text{ or } (15-√(335))/(10)`

Now we just need to find the corresponding y-coordinate for each pair of points.

I'm going to use the easier equation
`y=-3x+5`.

Let's do it for the first x I mentioned:

If
`x=(15+√(335))/(10)` then

`y=-3((15+√(335))/(10))+5`.

Let's simplify:

Distribute the -3 to the terms on top:

`y=(-45-3√(335))/(10)+5`

Combine the two terms; I'm going to do this by writing 5 as 50/10:

`y=(-45-3√(335)+50)/(10)`

Combine like terms on top; the -45+50 part:

`y=(5-3√(335))/(10)`.

So one solution point is:

`((15+√(335))/(10),(5-3√(335))/(10))`.

Let's find the other one for the other x that we got.

If
`x=(15-√(335))/(10)` then

`y=-3((15-√(335))/(10))+5`.

Let's simplify.

Distribute the -3 on top:

`y=(-45+3√(335))/(10)+5`

I'm going to write 5 as 50/10 so I can combine the terms as one fraction:

`y=(-45+3√(335)+50)/(5)`

Simplify the -45+50 part:

`y=(5+3√(335))/(10)`.

So the other point of intersection is:

`((15-√(335))/(10),(5+3√(335))/(10))`.

The two solutions in exact form are:

`((15+√(335))/(10),(5-3√(335))/(10))`

and

`((15-√(335))/(10),(5+3√(335))/(10))`.

If you prefer to look at approximations just put into your calculator:

`(3.3303,-4.9909)`

and

`(-0.3303,5.9909)`.