Solve {l}{4b-3a=92}{a+84=2b}

$$\left. \begin{array} { l } { 4 b - 3 a = 92 } \\ { a + 84 = 2 b } \end{array} \right.$$

Solve for b, a

$$b=80$$

$$a=76$$

Steps Using Substitution

Consider the second equation. Subtract 2b from both sides.

$$a+84-2b=0$$

Subtract 84 from both sides. Anything subtracted from zero gives its negation.

$$a-2b=-84$$

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.

$$4b-3a=92,-2b+a=-84$$

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

$$4b-3a=92$$

Add 3a to both sides of the equation.

$$4b=3a+92$$

Divide both sides by 4.

$$b=\frac{1}{4}\left(3a+92\right)$$

Multiply \frac{1}{4} times 3a+92.

$$b=\frac{3}{4}a+23$$

Substitute \frac{3a}{4}+23 for b in the other equation, -2b+a=-84.

$$-2\left(\frac{3}{4}a+23\right)+a=-84$$

Multiply -2 times \frac{3a}{4}+23.

$$-\frac{3}{2}a-46+a=-84$$

Add -\frac{3a}{2} to a.

$$-\frac{1}{2}a-46=-84$$

Add 46 to both sides of the equation.

$$-\frac{1}{2}a=-38$$

Multiply both sides by -2.

$$a=76$$

Substitute 76 for a in b=\frac{3}{4}a+23. Because the resulting equation contains only one variable, you can solve for b directly.

$$b=\frac{3}{4}\times 76+23$$

Multiply \frac{3}{4} times 76.

$$b=57+23$$

Add 23 to 57.

$$b=80$$

The system is now solved.

$$b=80,a=76$$

Steps Using Matrices

Consider the second equation. Subtract 2b from both sides.

$$a+84-2b=0$$

Subtract 84 from both sides. Anything subtracted from zero gives its negation.

$$a-2b=-84$$

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

$$4b-3a=92,-2b+a=-84$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}92\\-84\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-3\\-2&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right))\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right))\left(\begin{matrix}92\\-84\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right))\left(\begin{matrix}92\\-84\end{matrix}\right)$$

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

$$\left(\begin{matrix}b\\a\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\-2&1\end{matrix}\right))\left(\begin{matrix}92\\-84\end{matrix}\right)$$

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.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4-\left(-3\left(-2\right)\right)}&-\frac{-3}{4-\left(-3\left(-2\right)\right)}\\-\frac{-2}{4-\left(-3\left(-2\right)\right)}&\frac{4}{4-\left(-3\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}92\\-84\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&-\frac{3}{2}\\-1&-2\end{matrix}\right)\left(\begin{matrix}92\\-84\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 92-\frac{3}{2}\left(-84\right)\\-92-2\left(-84\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\a\end{matrix}\right)=\left(\begin{matrix}80\\76\end{matrix}\right)$$

Extract the matrix elements b and a.

$$b=80,a=76$$

Steps Using Elimination

Consider the second equation. Subtract 2b from both sides.

$$a+84-2b=0$$

Subtract 84 from both sides. Anything subtracted from zero gives its negation.

$$a-2b=-84$$

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.

$$4b-3a=92,-2b+a=-84$$

To make 4b and -2b equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 4.

$$-2\times 4b-2\left(-3\right)a=-2\times 92,4\left(-2\right)b+4a=4\left(-84\right)$$

Simplify.

$$-8b+6a=-184,-8b+4a=-336$$

Subtract -8b+4a=-336 from -8b+6a=-184 by subtracting like terms on each side of the equal sign.

$$-8b+8b+6a-4a=-184+336$$

Add -8b to 8b. Terms -8b and 8b cancel out, leaving an equation with only one variable that can be solved.

$$6a-4a=-184+336$$

Add 6a to -4a.

$$2a=-184+336$$

Add -184 to 336.

$$2a=152$$

Divide both sides by 2.

$$a=76$$

Substitute 76 for a in -2b+a=-84. Because the resulting equation contains only one variable, you can solve for b directly.

$$-2b+76=-84$$

Subtract 76 from both sides of the equation.

$$-2b=-160$$

Divide both sides by -2.

$$b=80$$

The system is now solved.

$$b=80,a=76$$