QuestionSolve for $x$ in the equation $gc=\frac{x+y}{-x}$ .

Studdy Solution

STEP 1

Assumptions1. The equation given is $xgc = \frac{x+y}{-x}$ , and we need to solve for $x$ . . $g$ and $c$ are constants.
3. $y$ is a variable, but we are not solving for it.
4. $x$ is not equal to zero.

STEP 2

First, we need to isolate $x$ on one side of the equation. To do this, we can multiply both sides of the equation by $-x$ to get rid of the denominator on the right side.
$-x \cdot xgc = -x \cdot \frac{x+y}{-x}$

STEP 3

implify the equation.
$-x^2gc = x+y$

STEP 4

Rearrange the equation to isolate $x$ terms on one side.
$-x^2gc - y = x$

STEP 5

Now, we need to factor out $x$ from the left side of the equation.
$x(-xgc -1) = y$

STEP 6

Finally, divide both sides of the equation by $-xgc -1$ to solve for $x$ .
$x = \frac{y}{-xgc -1}$ Therefore, the solution for $x$ is $x = \frac{y}{-xgc -1}$ .

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QuestionSolve for $x$ in the equation $gc=\frac{x+y}{-x}$ .

Studdy Solution

STEP 1

Assumptions1. The equation given is $xgc = \frac{x+y}{-x}$ , and we need to solve for $x$ . . $g$ and $c$ are constants.
3. $y$ is a variable, but we are not solving for it.
4. $x$ is not equal to zero.

STEP 2

First, we need to isolate $x$ on one side of the equation. To do this, we can multiply both sides of the equation by $-x$ to get rid of the denominator on the right side.
$-x \cdot xgc = -x \cdot \frac{x+y}{-x}$

STEP 3

implify the equation.
$-x^2gc = x+y$

STEP 4

Rearrange the equation to isolate $x$ terms on one side.
$-x^2gc - y = x$

STEP 5

Now, we need to factor out $x$ from the left side of the equation.
$x(-xgc -1) = y$

STEP 6

Finally, divide both sides of the equation by $-xgc -1$ to solve for $x$ .
$x = \frac{y}{-xgc -1}$ Therefore, the solution for $x$ is $x = \frac{y}{-xgc -1}$ .

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QuestionSolve for xxx in the equation gc=x+y−xgc=\frac{x+y}{-x}gc=−xx+y​.

Studdy Solution

STEP 1

Assumptions1. The equation given is xgc=x+y−xxgc = \frac{x+y}{-x}xgc=−xx+y​, and we need to solve for xxx. . ggg and ccc are constants.3. yyy is a variable, but we are not solving for it.4. xxx is not equal to zero.

STEP 2

First, we need to isolate xxx on one side of the equation. To do this, we can multiply both sides of the equation by −x-x−x to get rid of the denominator on the right side.−x⋅xgc=−x⋅x+y−x-x \cdot xgc = -x \cdot \frac{x+y}{-x}−x⋅xgc=−x⋅−xx+y​

STEP 3

implify the equation.−x2gc=x+y-x^2gc = x+y−x2gc=x+y

STEP 4

Rearrange the equation to isolate xxx terms on one side.−x2gc−y=x-x^2gc - y = x−x2gc−y=x

STEP 5

Now, we need to factor out xxx from the left side of the equation.x(−xgc−1)=yx(-xgc -1) = yx(−xgc−1)=y

STEP 6

Finally, divide both sides of the equation by −xgc−1-xgc -1−xgc−1 to solve for xxx.x=y−xgc−1x = \frac{y}{-xgc -1}x=−xgc−1y​Therefore, the solution for xxx is x=y−xgc−1x = \frac{y}{-xgc -1}x=−xgc−1y​.

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QuestionSolve for $x$ in the equation $gc=\frac{x+y}{-x}$ .

Assumptions1. The equation given is $xgc = \frac{x+y}{-x}$ , and we need to solve for $x$ . . $g$ and $c$ are constants.
3. $y$ is a variable, but we are not solving for it.
4. $x$ is not equal to zero.

First, we need to isolate $x$ on one side of the equation. To do this, we can multiply both sides of the equation by $-x$ to get rid of the denominator on the right side.
$-x \cdot xgc = -x \cdot \frac{x+y}{-x}$

implify the equation.
$-x^2gc = x+y$

Rearrange the equation to isolate $x$ terms on one side.
$-x^2gc - y = x$

Now, we need to factor out $x$ from the left side of the equation.
$x(-xgc -1) = y$

Finally, divide both sides of the equation by $-xgc -1$ to solve for $x$ .
$x = \frac{y}{-xgc -1}$ Therefore, the solution for $x$ is $x = \frac{y}{-xgc -1}$ .