Question

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

Answer 1

Answer:

The two solutions in exact form are:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$

and

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$.

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

$(3.3303,-4.9909)$

and

$(-0.3303,5.9909)$.

Step-by-step 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:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

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

$x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{2(10)}$

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

$x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{20}$

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

$x=\frac{30 \pm \sqrt{1340}}{20}$

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:

$\sqrt{1340}=\sqrt{4}\sqrt{335}=2\sqrt{335}$.

Let's go back to the solution:

$x=\frac{30 \pm \sqrt{1340}}{20}$

$x=\frac{30 \pm 2 \sqrt{335}}{20}$

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

$x=\frac{\frac{30}{2} \pm \frac{2 \sqrt{335}}{2}}{\frac{20}{2}}$

$x=\frac{15 \pm \sqrt{335}}{10}$

So we have these two x values:

$x=\frac{15+\sqrt{335}}{10} \text{ or } \frac{15-\sqrt{335}}{10}$

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 $x=\frac{15+\sqrt{335}}{10}$ then

$y=-3(\frac{15+\sqrt{335}}{10})+5$.

Let's simplify:

Distribute the -3 to the terms on top:

$y=\frac{-45-3\sqrt{335}}{10}+5$

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

$y=\frac{-45-3\sqrt{335}+50}{10}$

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

$y=\frac{5-3\sqrt{335}}{10}$.

So one solution point is:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$.

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

If $x=\frac{15-\sqrt{335}}{10}$ then

$y=-3(\frac{15-\sqrt{335}}{10})+5$.

Let's simplify.

Distribute the -3 on top:

$y=\frac{-45+3\sqrt{335}}{10}+5$

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

$y=\frac{-45+3\sqrt{335}+50}{5}$

Simplify the -45+50 part:

$y=\frac{5+3\sqrt{335}}{10}$.

So the other point of intersection is:

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$.

The two solutions in exact form are:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$

and

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$.

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

$(3.3303,-4.9909)$

and

$(-0.3303,5.9909)$.