Solve {l}{2lambda-3b=-12}{5lambda-8b=-40}

$$\left. \begin{array} { l } { 2 \lambda - 3 b = - 12 } \\ { 5 \lambda - 8 b = - 40 } \end{array} \right.$$

Solve for λ, b

$$\lambda =24$$

$$b=20$$

Steps Using Substitution

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.

$$2\lambda -3b=-12,5\lambda -8b=-40$$

Choose one of the equations and solve it for \lambda by isolating \lambda on the left hand side of the equal sign.

$$2\lambda -3b=-12$$

Add 3b to both sides of the equation.

$$2\lambda =3b-12$$

Divide both sides by 2.

$$\lambda =\frac{1}{2}\left(3b-12\right)$$

Multiply \frac{1}{2} times -12+3b.

$$\lambda =\frac{3}{2}b-6$$

Substitute \frac{3b}{2}-6 for \lambda in the other equation, 5\lambda -8b=-40.

$$5\left(\frac{3}{2}b-6\right)-8b=-40$$

Multiply 5 times \frac{3b}{2}-6.

$$\frac{15}{2}b-30-8b=-40$$

Add \frac{15b}{2} to -8b.

$$-\frac{1}{2}b-30=-40$$

Add 30 to both sides of the equation.

$$-\frac{1}{2}b=-10$$

Multiply both sides by -2.

$$b=20$$

Substitute 20 for b in \lambda =\frac{3}{2}b-6. Because the resulting equation contains only one variable, you can solve for \lambda directly.

$$\lambda =\frac{3}{2}\times 20-6$$

Multiply \frac{3}{2} times 20.

$$\lambda =30-6$$

Add -6 to 30.

$$\lambda =24$$

The system is now solved.

$$\lambda =24,b=20$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$2\lambda -3b=-12,5\lambda -8b=-40$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right)\left(\begin{matrix}\lambda \\b\end{matrix}\right)=\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\5&-8\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right))\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right)\left(\begin{matrix}\lambda \\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right))\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}\lambda \\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right))\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}\lambda \\b\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\5&-8\end{matrix}\right))\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}\lambda \\b\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{2\left(-8\right)-\left(-3\times 5\right)}&-\frac{-3}{2\left(-8\right)-\left(-3\times 5\right)}\\-\frac{5}{2\left(-8\right)-\left(-3\times 5\right)}&\frac{2}{2\left(-8\right)-\left(-3\times 5\right)}\end{matrix}\right)\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}\lambda \\b\end{matrix}\right)=\left(\begin{matrix}8&-3\\5&-2\end{matrix}\right)\left(\begin{matrix}-12\\-40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}\lambda \\b\end{matrix}\right)=\left(\begin{matrix}8\left(-12\right)-3\left(-40\right)\\5\left(-12\right)-2\left(-40\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}\lambda \\b\end{matrix}\right)=\left(\begin{matrix}24\\20\end{matrix}\right)$$

Extract the matrix elements \lambda and b.

$$\lambda =24,b=20$$

Steps Using Elimination

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.

$$2\lambda -3b=-12,5\lambda -8b=-40$$

To make 2\lambda and 5\lambda equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 2.

$$5\times 2\lambda +5\left(-3\right)b=5\left(-12\right),2\times 5\lambda +2\left(-8\right)b=2\left(-40\right)$$

Simplify.

$$10\lambda -15b=-60,10\lambda -16b=-80$$

Subtract 10\lambda -16b=-80 from 10\lambda -15b=-60 by subtracting like terms on each side of the equal sign.

$$10\lambda -10\lambda -15b+16b=-60+80$$

Add 10\lambda to -10\lambda . Terms 10\lambda and -10\lambda cancel out, leaving an equation with only one variable that can be solved.

$$-15b+16b=-60+80$$

Add -15b to 16b.

$$b=-60+80$$

Add -60 to 80.

$$b=20$$

Substitute 20 for b in 5\lambda -8b=-40. Because the resulting equation contains only one variable, you can solve for \lambda directly.

$$5\lambda -8\times 20=-40$$

Multiply -8 times 20.

$$5\lambda -160=-40$$

Add 160 to both sides of the equation.

$$5\lambda =120$$

Divide both sides by 5.

$$\lambda =24$$

The system is now solved.

$$\lambda =24,b=20$$