For each equation, (b) solve for y in terms of x. See Example ...

Question

For each equation, (b) solve for y in terms of x. See Example 8.

$2 x^{2} + 4 x y - 3 y^{2} = 2$

Accepted Answer

Hey, everyone here, we are asked to express Y in terms of X for the following equation, nine X squared minus 15 X Y plus five Y squared is equal to seven. Now, our first step here is to move all of our terms to the same side of the equation. So we just need to move positive seven from the right side to the left side by subtracting seven from both sides of our equation. So now we have nine X squared minus 15 X Y plus five Y squared minus seven is equal to zero. So now we just want to rearrange the terms in this equation. So first, we will have five Y squared and then we will have our other Y term which is minus 15 X Y, then we'll have our other two terms. So we'll have plus nine X squared minus seven. And then we have this is all equal to zero. So now we see that we can, we actually have the quad, a quadratic equation with the variable Y. So if we recall the formula for a quadratic equation, we have a Y squared plus B Y plus C is equal to zero. So now we just want to identify our values for A B and C by comparing this formula to our equation. So we can start with the coefficient of Y squared, which is the value for A. So A is just going to be five and next our value for B is going to be the coefficient of Y. So B is just going to be negative X and lastly C is going to be the rest of our terms here. And we see that these terms do not contain the variable Y. So C is going to be nine X squared minus seven. And so now with these values for A B and C identified, we can use the quadratic formula to rewrite our original equation as Y in terms of X. So if we recall the quadratic formula, we know that Y is equal to negative B plus or minus the square root of B squared minus four A C all divided by two A. And so now we just need to substitute in our values for A B and C to rewrite our equation equation as why? So doing this, we have Y is equal to negative quantity of negative 15 X plus or minus the square root of the quantity of negative 15 X squared minus four times five times the quantity of nine X squared minus seven. And again, this is all divided by two times A which is in our case, two times five. And so now we just need to simplify the right side of our equation. So we have positive 15 X plus or minus the square root of. And now negative 15 X squared is and 25 X squared. And then we have minus 20 times the quantity of nine X squared minus seven all divided by two times five, which is 10. And now continuing to simplify, we have 15 X plus or minus the square root of 225 X squared minus 180 X squared plus 140. And again, this is all divided by 10. And now continuing to simplify this expression, we just need to simplify the route a bit further. So we have Y is equal to 15 X plus or minus the square root of now 225 minus 180 gives us 45 X squared plus 140. And again, this is all divided by 10. And with this, we have rewritten our original equation as Y in terms of X by applying the quadratic formula. So this final equation is our final answer which leaves us with answer. Choice. B thanks for watching. I hope you found this video helpful and I'll see you again next time.

For each equation, (b) solve for y in terms of x. See Example 8.
$2 x^{2} + 4 x y - 3 y^{2} = 2$

Key Concepts

Solving Quadratic Equations in Terms of One Variable

Quadratic Formula

Algebraic Manipulation and Rearrangement

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