Skip to main content
Solve for k, p
Tick mark Image

Similar Problems from Web Search

Share

2.5k+p=120,3k+p=80
To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
2.5k+p=120
Choose one of the equations and solve it for k by isolating k on the left hand side of the equal sign.
2.5k=-p+120
Subtract p from both sides of the equation.
k=0.4\left(-p+120\right)
Divide both sides of the equation by 2.5, which is the same as multiplying both sides by the reciprocal of the fraction.
k=-0.4p+48
Multiply 0.4 times -p+120.
3\left(-0.4p+48\right)+p=80
Substitute -\frac{2p}{5}+48 for k in the other equation, 3k+p=80.
-1.2p+144+p=80
Multiply 3 times -\frac{2p}{5}+48.
-0.2p+144=80
Add -\frac{6p}{5} to p.
-0.2p=-64
Subtract 144 from both sides of the equation.
p=320
Multiply both sides by -5.
k=-0.4\times 320+48
Substitute 320 for p in k=-0.4p+48. Because the resulting equation contains only one variable, you can solve for k directly.
k=-128+48
Multiply -0.4 times 320.
k=-80
Add 48 to -128.
k=-80,p=320
The system is now solved.
2.5k+p=120,3k+p=80
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right)\left(\begin{matrix}k\\p\end{matrix}\right)=\left(\begin{matrix}120\\80\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right))\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right)\left(\begin{matrix}k\\p\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right))\left(\begin{matrix}120\\80\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2.5&1\\3&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k\\p\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right))\left(\begin{matrix}120\\80\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}k\\p\end{matrix}\right)=inverse(\left(\begin{matrix}2.5&1\\3&1\end{matrix}\right))\left(\begin{matrix}120\\80\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}k\\p\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2.5-3}&-\frac{1}{2.5-3}\\-\frac{3}{2.5-3}&\frac{2.5}{2.5-3}\end{matrix}\right)\left(\begin{matrix}120\\80\end{matrix}\right)
For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.
\left(\begin{matrix}k\\p\end{matrix}\right)=\left(\begin{matrix}-2&2\\6&-5\end{matrix}\right)\left(\begin{matrix}120\\80\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}k\\p\end{matrix}\right)=\left(\begin{matrix}-2\times 120+2\times 80\\6\times 120-5\times 80\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}k\\p\end{matrix}\right)=\left(\begin{matrix}-80\\320\end{matrix}\right)
Do the arithmetic.
k=-80,p=320
Extract the matrix elements k and p.
2.5k+p=120,3k+p=80
In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.
2.5k-3k+p-p=120-80
Subtract 3k+p=80 from 2.5k+p=120 by subtracting like terms on each side of the equal sign.
2.5k-3k=120-80
Add p to -p. Terms p and -p cancel out, leaving an equation with only one variable that can be solved.
-0.5k=120-80
Add \frac{5k}{2} to -3k.
-0.5k=40
Add 120 to -80.
k=-80
Multiply both sides by -2.
3\left(-80\right)+p=80
Substitute -80 for k in 3k+p=80. Because the resulting equation contains only one variable, you can solve for p directly.
-240+p=80
Multiply 3 times -80.
p=320
Add 240 to both sides of the equation.
k=-80,p=320
The system is now solved.