Solve for x in terms of y?
You said it correctly the first time - you want to solve for x in terms of y. You need to "complete the square", then take the square root: y = -3 + 2x + x^2 = (x^2 + 2x + 1) - 4 = (x+1)^2 - 4 y+4 = (x+1)^2 x+1 = +/-sqrt(y+4) x = -1 +/- sqrt(y+4)
Solve for x in terms of y: y = 2x/(x-2)?
Hi, y = 2x/(x-2) Your first step was correct. To get rid of the fraction multiply both sides by the denominator, (x - 2) y( x - 2) = 2x Next, get rid of the parentheses by using the Distributive Property: xy - 2y = 2x Since you want to solve for x, you must move all terms with x to one side of the equation and all terms without x to the other side. That means to subtract xy. -2y = 2x - xy In order to solve for x, you only want it to appear once it your equation, so factor it out as a common factor on the right side. If anything else could also be factored out with the x, don't take them - just the x! -2y = x(2 - y) Now to finish getting the x alone, divide both sides by what is multiplying it, the expression 2 - y. ...-2y.........x(2 - y) --------..=..----------- On the right, the 2 - y factors cancel out. ..2.-.y..........2..-.y ...-2y --------..=..x This is your answer! ..2.-.y I hope that helps.
How to solve for y in terms of x?
isolate for y... you pretty much have to do the opposite operation each side if you want to do this.. soo.. 1) 2x-3y = 12 -3y = 12 - 2x y = -12/3 + 2x/3 y = -4 + 2/3x or y = (2/3)x - 4 2) 5x+2y = 7 2y = 7 - 5x y = 7/2 - 5/2x or y = (7/2) - (5/2)x so for the first one, you first subtract 2x from both sides. Then divide both sides by -3 and simplify. For the second one, you first subtract 5x from both sides, then divide both sides by 2.
How can I solve for y in terms of x? X+y=5?
solving for y in terms of x is the same as saying y depends on x.So, if something happens to x, it has an impact on y. In linear equations like these and others that you will find as you move forward in math, the output is by itself on one side and the input will be on the other side among other numbers and variables( inputs). This means we need y on one side by itself and x on the other side. Well, right now x and y are on the same side, but we need y by itself because our problem is telling us that y depends on x OR y is our output when x is the input. Lets move x to the other side. How do you get rid of x? You subtract it away. So subtract x from the left side. This will look like x-x+y = 5. We know x-x = 0 so you are left with y = 5, BUT WAIT! This looks different. We can't just take away something from one side and not the other. Remember this: if you do something to one side you must do it to the other! You take away x from one side then also take away x from the other. Ok cool so y = 5-x. As you can see, you input is on the right side with a 5 and your output is on the left all by itself. You input a value for x = 1, you get an output y = 4. Notice how this is an equation of a line. Can you tell what the slope is of this line?
Solve for x and y in terms of a and b.?
5x + 2y = a ... eqn 1 x - y = b ... eqn 2 eqn 2 is simpler than eqn 1 so we'll use eqn 2 eqn 2: x - y = b thus x = b + y ... eqn 3 sub eqn 3 into eqn 1 eqn 1: 5x + 2y = a after sub eqn 1 becomes: 5(b + y) + 2y = a thus: 5b + 5y + 2y = a 7y = a - 5b y = (a - 5b)/7 sub y into eqn 3 x = b + y thus x = b + (a - 5b)/7
How can I solve for y in terms of x?
We have [math]\frac{e(iz) + e(-iz)}{2}= cos(z)[/math]So if you let [math]z = \frac{y}{i}[/math]Then you get [math]\frac{e(y) + e(-y)}{2} = cos(\frac{y}{i})[/math]More over [math]\frac{e(y) + e(-y)}{2} = x[/math]Hence [math]cos(\frac{y}{i}) = x[/math]Then [math]y = i * arcos(x)[/math]
Can you solve [math]x-xy=1-x[/math] in terms of [math]x[/math]?
Yes, solve for “x” so that the equation resembles: “x=…”When you bring all of the “x” terms to the left, this is the equation:2x-xy=1Factoring “x” extracts the “x” factor from the term containing “y”:x(2-y)=1Now me must divide each side of the equation by the “(2-y)” factor; this produces the solved equation:x= 1/(2-y)Some professors and textbooks prefer non-negative variables, so as a last touch, you can factor the negative from the denominator and express the equation as:x=(-1)/(y-2)Your instructor may (or may not) require this additional factoring step.
How do you solve for x in in terms of y in [math]x+y = xy[/math]?
[math]x + y = xy[/math]Subtract [math]x[/math] from both sides.[math]y = xy - x[/math]Factor out the common factor of [math]x[/math].[math]y = x(y-1)[/math]Divide to put [math]x[/math] by itself.[math]x = \dfrac{y}{y-1}[/math]If you had tried subtracting y from both sides first, you would find that you had solved for y, therefore you would be able to figure out that you needed to subtract x from both sides to solve for x. In all honesty, that is what I did first :)You notice that by dividing by [math]y-1[/math] means [math]y \ne 1[/math], but if you plug [math]y=1[/math] into the original equation, that was true there also.[math]x + 1 = x(1)[/math][math]x + 1 \ne x[/math]Therefore, the algebraic manipulation did not change the domain.
How do you solve for X and y in terms of u and v when u = xy and v = xy^3?
"Suppose that x and y are two independent variables which can be expressed in terms of two other independent variables u and v by the formula x = g(u,v) and y = h(u,v). The Jacobian of x and y with respect to u and v, J(u,v), isJ(u,v) = | (dx/du) (dx/dv) |Det | (dy/du) (dy/dv) |= (dx/du)(dy/dv) - (dy/du)(dx/dv)"where d represents a partial derivative.From this, all we need to do to continue is swap: x,y <-> u,v. This reduces the need for quotient rule and/or nasty expressions using non-elementary functions to express x or y strictly in terms of u and v.Thus, x = uv and y = uv^3.dx/du = vdx/dv = udy/du = v^3dy/dv = 3uv^2So J(u,v) = (v)(3uv^2) - (u)(v^3) = 3uv^3 - uv^3 = 2uv^3.Cited source: https://math.kennesaw.edu/~plava...
What does it mean to solve for y in terms of x?
Basically, you’re trying to get to a statement that has y = where you only see the y by itself on one side of the equals sign.You can think about the two sides of the equals sign as being separated by a net.Imagine you have the left and right side and there is a y on the left side and some other things and there are some other things (but not y) on the right side. I’ll use squares, stars, triangles, and circles.You would then want to remove things from the left side by subtracting. If you subtract something from itself, you’re left with zero of that thing.In a more typical algebra equation, trying to solve for yAgain, you’re trying to isolate y =So you can bring the terms with y either all to the left side or all to the right side. I brought them all to the left side. And I subtracted things from the left side that did not contain y. Since the two sides are equal (that’s what the equals sign means) then if you change the left side, you have to change the right side in the same way.You can also use multiplication, division, and other things depending on what you start with.Here, you have 2 y, and you want just y = , so you want to divide both sides by 2.Physics and Math Questions?