Math / Algebra

Questionpoints Points: 0 of 1
For the equation, (a) solve for $x$ in terms of $y$ , and (b) solve for $y$ in terms of $x$ . $8 x^{2}+2 x y+5 y^{2}=8$ (a) Solve for $x$ in terms of $y$ . $x=\square$ (Simplify your answer, including any radicals. Use integers or fractions for any numbers in the expressi. comma to separate answers as needed.)

Studdy Solution

STEP 1

What is this asking? We've got this cool equation with x and y all mixed up, and we want to untangle it so x is on one side and everything else is on the other, and then do the same for y! Watch out! Remember the quadratic formula?
We might need that, and don't forget those plus-or-minus signs – they're sneaky little things!

STEP 2

1. Solve for x
2. Solve for y

STEP 3

Alright, let's rewrite our equation: $8x^2 + 2xy + 5y^2 = 8$ We want to solve for $x$ , so let's rearrange this to look like a quadratic equation in terms of $x$ .
Think of $y$ as just another number for now!

STEP 4

Rearrange the equation: $8x^2 + (2y)x + (5y^2 - 8) = 0$ See how we grouped the terms?
Now it looks just like $ax^2 + bx + c = 0$ , where $a = 8$ , $b = 2y$ , and $c = 5y^2 - 8$ .

STEP 5

Time for the quadratic formula!
Remember, it's $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 6

Let's plug in our a, b, and c: $x = \frac{-(2y) \pm \sqrt{(2y)^2 - 4 \cdot 8 \cdot (5y^2 - 8)}}{2 \cdot 8}$

STEP 7

Simplify that monster: $x = \frac{-2y \pm \sqrt{4y^2 - 160y^2 + 256}}{16}$ $x = \frac{-2y \pm \sqrt{-156y^2 + 256}}{16}$ $x = \frac{-2y \pm \sqrt{4(-39y^2 + 64)}}{16}$ $x = \frac{-2y \pm 2\sqrt{-39y^2 + 64}}{16}$ $x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}$ Boom! We solved for x in terms of y.

STEP 8

Now, let's solve for $y$ .
We start with the same equation: $8x^2 + 2xy + 5y^2 = 8$

STEP 9

Rewrite it to look like a quadratic equation in terms of $y$ : $5y^2 + (2x)y + (8x^2 - 8) = 0$ Here, $a = 5$ , $b = 2x$ , and $c = 8x^2 - 8$ .

STEP 10

Quadratic formula time again! $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 11

Plug in our new a, b, and c: $y = \frac{-(2x) \pm \sqrt{(2x)^2 - 4 \cdot 5 \cdot (8x^2 - 8)}}{2 \cdot 5}$

STEP 12

Simplify: $y = \frac{-2x \pm \sqrt{4x^2 - 160x^2 + 160}}{10}$ $y = \frac{-2x \pm \sqrt{-156x^2 + 160}}{10}$ $y = \frac{-2x \pm \sqrt{4(-39x^2 + 40)}}{10}$ $y = \frac{-2x \pm 2\sqrt{40 - 39x^2}}{10}$ $y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}$ And there we have it, y in terms of x!

STEP 13

(a) $x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}$ (b) $y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}$

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Studdy Solution

STEP 1

STEP 2

1. Solve for x
2. Solve for y

STEP 3

Alright, let's rewrite our equation: $8x^2 + 2xy + 5y^2 = 8$ We want to solve for $x$ , so let's rearrange this to look like a quadratic equation in terms of $x$ .
Think of $y$ as just another number for now!

STEP 4

Rearrange the equation: $8x^2 + (2y)x + (5y^2 - 8) = 0$ See how we grouped the terms?
Now it looks just like $ax^2 + bx + c = 0$ , where $a = 8$ , $b = 2y$ , and $c = 5y^2 - 8$ .

STEP 5

Time for the quadratic formula!
Remember, it's $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 6

Let's plug in our a, b, and c: $x = \frac{-(2y) \pm \sqrt{(2y)^2 - 4 \cdot 8 \cdot (5y^2 - 8)}}{2 \cdot 8}$

STEP 7

Simplify that monster: $x = \frac{-2y \pm \sqrt{4y^2 - 160y^2 + 256}}{16}$ $x = \frac{-2y \pm \sqrt{-156y^2 + 256}}{16}$ $x = \frac{-2y \pm \sqrt{4(-39y^2 + 64)}}{16}$ $x = \frac{-2y \pm 2\sqrt{-39y^2 + 64}}{16}$ $x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}$ Boom! We solved for x in terms of y.

STEP 8

Now, let's solve for $y$ .
We start with the same equation: $8x^2 + 2xy + 5y^2 = 8$

STEP 9

Rewrite it to look like a quadratic equation in terms of $y$ : $5y^2 + (2x)y + (8x^2 - 8) = 0$ Here, $a = 5$ , $b = 2x$ , and $c = 8x^2 - 8$ .

STEP 10

Quadratic formula time again! $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

STEP 11

Plug in our new a, b, and c: $y = \frac{-(2x) \pm \sqrt{(2x)^2 - 4 \cdot 5 \cdot (8x^2 - 8)}}{2 \cdot 5}$

STEP 12

Simplify: $y = \frac{-2x \pm \sqrt{4x^2 - 160x^2 + 160}}{10}$ $y = \frac{-2x \pm \sqrt{-156x^2 + 160}}{10}$ $y = \frac{-2x \pm \sqrt{4(-39x^2 + 40)}}{10}$ $y = \frac{-2x \pm 2\sqrt{40 - 39x^2}}{10}$ $y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}$ And there we have it, y in terms of x!

STEP 13

(a) $x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}$ (b) $y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}$

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Studdy Solution

STEP 1

STEP 2

1. Solve for *x*2. Solve for *y*

STEP 3

Alright, let's **rewrite** our equation: 8x2+2xy+5y2=88x^2 + 2xy + 5y^2 = 88x2+2xy+5y2=8 We want to solve for xxx, so let's rearrange this to look like a quadratic equation in terms of xxx.Think of yyy as just another number for now!

STEP 4

**Rearrange** the equation: 8x2+(2y)x+(5y2−8)=08x^2 + (2y)x + (5y^2 - 8) = 08x2+(2y)x+(5y2−8)=0 See how we grouped the terms?Now it looks just like ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, where a=8a = 8a=8, b=2yb = 2yb=2y, and c=5y2−8c = 5y^2 - 8c=5y2−8.

STEP 5

Time for the **quadratic formula**!Remember, it's x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}x=2a−b±b2−4ac​​

STEP 6

Let's **plug in** our *a*, *b*, and *c*: x=−(2y)±(2y)2−4⋅8⋅(5y2−8)2⋅8x = \frac{-(2y) \pm \sqrt{(2y)^2 - 4 \cdot 8 \cdot (5y^2 - 8)}}{2 \cdot 8}x=2⋅8−(2y)±(2y)2−4⋅8⋅(5y2−8)​​

STEP 7

STEP 8

Now, let's **solve for** yyy.We start with the same equation: 8x2+2xy+5y2=88x^2 + 2xy + 5y^2 = 88x2+2xy+5y2=8

STEP 9

**Rewrite** it to look like a quadratic equation in terms of yyy: 5y2+(2x)y+(8x2−8)=05y^2 + (2x)y + (8x^2 - 8) = 05y2+(2x)y+(8x2−8)=0 Here, a=5a = 5a=5, b=2xb = 2xb=2x, and c=8x2−8c = 8x^2 - 8c=8x2−8.

STEP 10

**Quadratic formula** time again! y=−b±b2−4ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}y=2a−b±b2−4ac​​

STEP 11

**Plug in** our new *a*, *b*, and *c*: y=−(2x)±(2x)2−4⋅5⋅(8x2−8)2⋅5y = \frac{-(2x) \pm \sqrt{(2x)^2 - 4 \cdot 5 \cdot (8x^2 - 8)}}{2 \cdot 5}y=2⋅5−(2x)±(2x)2−4⋅5⋅(8x2−8)​​

STEP 12

STEP 13

(a) x=−y±64−39y28x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}x=8−y±64−39y2​​ (b) y=−x±40−39x25y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}y=5−x±40−39x2​​

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1. Solve for x
2. Solve for y

Alright, let's rewrite our equation: $8x^2 + 2xy + 5y^2 = 8$ We want to solve for $x$ , so let's rearrange this to look like a quadratic equation in terms of $x$ .
Think of $y$ as just another number for now!

Rearrange the equation: $8x^2 + (2y)x + (5y^2 - 8) = 0$ See how we grouped the terms?
Now it looks just like $ax^2 + bx + c = 0$ , where $a = 8$ , $b = 2y$ , and $c = 5y^2 - 8$ .

Time for the quadratic formula!
Remember, it's $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Let's plug in our a, b, and c: $x = \frac{-(2y) \pm \sqrt{(2y)^2 - 4 \cdot 8 \cdot (5y^2 - 8)}}{2 \cdot 8}$

Now, let's solve for $y$ .
We start with the same equation: $8x^2 + 2xy + 5y^2 = 8$

Rewrite it to look like a quadratic equation in terms of $y$ : $5y^2 + (2x)y + (8x^2 - 8) = 0$ Here, $a = 5$ , $b = 2x$ , and $c = 8x^2 - 8$ .

Quadratic formula time again! $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Plug in our new a, b, and c: $y = \frac{-(2x) \pm \sqrt{(2x)^2 - 4 \cdot 5 \cdot (8x^2 - 8)}}{2 \cdot 5}$

(a) $x = \frac{-y \pm \sqrt{64 - 39y^2}}{8}$ (b) $y = \frac{-x \pm \sqrt{40 - 39x^2}}{5}$