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.

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

Accepted Answer

Hey, everyone here, we are asked to express Y in terms of X for the following equation, 13 X squared plus eight X Y minus 10, Y squared is equal to six. Now, our first step here is to set our equation equal to zero. So that means we need to move this positive six from the right side to the left side of our equation. So we have 13 X squared plus eight X Y minus 10, Y squared minus six is equal to zero. So now we just want to rearrange our equation that we have so far so that our Y terms are in the beginning of the equation on the left hand side. So doing this, we have negative 10 Y squared plus eight X Y plus 13 X squared minus six is equal to zero. So now looking at our equation, we see that it matches the form of a quadratic equation which takes the form of 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 and we need to note that our variable X will be treated as a constant since we are solving for the variable Y. So doing this, we can start by finding our value for a. Well, A is just the coefficient of Y squared. So A is going to be negative 10 and now moving on to B well B is the coefficient of Y. So B is going to be A X. And next our remaining values are our constants. So our value for C is going to to be these values which are 13 X squared minus six. So now with these values identified, we can apply our quadratic formula. But first, we need to recall the quadratic formula which tells us that why is equal to negative B plus or minus the square root of B squared minus four A C all divided by two A. So again, we have our A B and C values identified. So we can just substitute them into our formula here. And we see that Y is equal to negative A X plus or minus the square root of the quantity of eight X squared minus minus four times negative 10 times the quantity of 13 X squared minus six. And this is all divided by two times A which is just two times negative 10. So now we just need to continue to simplify this expression. So we have Y is equal to negative eight X plus or minus the square root of eight X squared gives us 64 X squared. Plus 40 times the quantity of 13 X squared minus six all divided by two times negative 10, which is just negative 20. So now we can continue to simplify. First, we see that we have a negative in the numerator and the denominator. So they will cancel and become positive. And we also want to factor in this 40 that's underneath the root. So doing this again, we have now positive eight X plus or minus the square root of 64 X squared plus 40 times 13 gives us 512 X squared. And then we have minus 240 this is all divided by positive 20. And now we just have to combine the like terms underneath the root and we find that Y is equal to eight X plus or minus the square root of 64 plus 520 gives us 584 X squared minus 240 All divided by positive 20. And so now we have finished solving for Y in terms of X. So this equation is our final solution here which leaves us with answer choice. A 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.
$4 x^{2} - 2 x y + 3 y^{2} = 2$

Key Concepts

Solving for a Variable

Quadratic Equations in Two Variables

Using the Quadratic Formula

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