Solve find the value of ) 9k ^ 2 - kx + 1b = 0 is a a perfect | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$\left. \begin{array} { l } { 9 k ^ { 2 } - k x + 16 = 0 } \end{array} \right.$$

Solve for x

$x=9k+\frac{16}{k}$
$k\neq 0$

Steps for Solving Linear Equation

Subtract $9k^{2}$ from both sides. Anything subtracted from zero gives its negation.

$$-kx+16=-9k^{2}$$

Subtract $16$ from both sides.

$$-kx=-9k^{2}-16$$

The equation is in standard form.

$$\left(-k\right)x=-9k^{2}-16$$

Divide both sides by $-k$.

$$\frac{\left(-k\right)x}{-k}=\frac{-9k^{2}-16}{-k}$$

Dividing by $-k$ undoes the multiplication by $-k$.

$$x=\frac{-9k^{2}-16}{-k}$$

Divide $-9k^{2}-16$ by $-k$.

$$x=9k+\frac{16}{k}$$