Solve for k
k=\frac{5}{2\left(1-3p\right)}
p\neq \frac{1}{3}
Solve for p
p=\frac{1}{3}-\frac{5}{6k}
k\neq 0
Share
Copied to clipboard
2k\left(-3p+1\right)=5
Multiply both sides of the equation by -3p+1.
-6kp+2k=5
Use the distributive property to multiply 2k by -3p+1.
\left(-6p+2\right)k=5
Combine all terms containing k.
\left(2-6p\right)k=5
The equation is in standard form.
\frac{\left(2-6p\right)k}{2-6p}=\frac{5}{2-6p}
Divide both sides by -6p+2.
k=\frac{5}{2-6p}
Dividing by -6p+2 undoes the multiplication by -6p+2.
k=\frac{5}{2\left(1-3p\right)}
Divide 5 by -6p+2.
2k\left(-3p+1\right)=5
Variable p cannot be equal to \frac{1}{3} since division by zero is not defined. Multiply both sides of the equation by -3p+1.
-6kp+2k=5
Use the distributive property to multiply 2k by -3p+1.
-6kp=5-2k
Subtract 2k from both sides.
\left(-6k\right)p=5-2k
The equation is in standard form.
\frac{\left(-6k\right)p}{-6k}=\frac{5-2k}{-6k}
Divide both sides by -6k.
p=\frac{5-2k}{-6k}
Dividing by -6k undoes the multiplication by -6k.
p=\frac{1}{3}-\frac{5}{6k}
Divide 5-2k by -6k.
p=\frac{1}{3}-\frac{5}{6k}\text{, }p\neq \frac{1}{3}
Variable p cannot be equal to \frac{1}{3}.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}