Solve {l}{x-29z=15}{4x+3z=-2}

$$\left. \begin{array} { l } { x - 29 z = 15 } \\ { 4 x + 3 z = - 2 } \end{array} \right.$$

Solve for x, z

$$x=-\frac{13}{119}\approx -0.109243697$$

$$z=-\frac{62}{119}\approx -0.521008403$$

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.

$$x-29z=15,4x+3z=-2$$

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

$$x-29z=15$$

Add 29z to both sides of the equation.

$$x=29z+15$$

Substitute 29z+15 for x in the other equation, 4x+3z=-2.

$$4\left(29z+15\right)+3z=-2$$

Multiply 4 times 29z+15.

$$116z+60+3z=-2$$

Add 116z to 3z.

$$119z+60=-2$$

Subtract 60 from both sides of the equation.

$$119z=-62$$

Divide both sides by 119.

$$z=-\frac{62}{119}$$

Substitute -\frac{62}{119} for z in x=29z+15. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=29\left(-\frac{62}{119}\right)+15$$

Multiply 29 times -\frac{62}{119}.

$$x=-\frac{1798}{119}+15$$

Add 15 to -\frac{1798}{119}.

$$x=-\frac{13}{119}$$

The system is now solved.

$$x=-\frac{13}{119},z=-\frac{62}{119}$$

Steps Using Matrices

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

$$x-29z=15,4x+3z=-2$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-29\\4&3\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}15\\-2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-29\\4&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-29\\4&3\end{matrix}\right))\left(\begin{matrix}1&-29\\4&3\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}1&-29\\4&3\end{matrix}\right))\left(\begin{matrix}15\\-2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}1&-29\\4&3\end{matrix}\right))\left(\begin{matrix}15\\-2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}1&-29\\4&3\end{matrix}\right))\left(\begin{matrix}15\\-2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3-\left(-29\times 4\right)}&-\frac{-29}{3-\left(-29\times 4\right)}\\-\frac{4}{3-\left(-29\times 4\right)}&\frac{1}{3-\left(-29\times 4\right)}\end{matrix}\right)\left(\begin{matrix}15\\-2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}\frac{3}{119}&\frac{29}{119}\\-\frac{4}{119}&\frac{1}{119}\end{matrix}\right)\left(\begin{matrix}15\\-2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}\frac{3}{119}\times 15+\frac{29}{119}\left(-2\right)\\-\frac{4}{119}\times 15+\frac{1}{119}\left(-2\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{119}\\-\frac{62}{119}\end{matrix}\right)$$

Extract the matrix elements x and z.

$$x=-\frac{13}{119},z=-\frac{62}{119}$$

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.

$$x-29z=15,4x+3z=-2$$

To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.

$$4x+4\left(-29\right)z=4\times 15,4x+3z=-2$$

Simplify.

$$4x-116z=60,4x+3z=-2$$

Subtract 4x+3z=-2 from 4x-116z=60 by subtracting like terms on each side of the equal sign.

$$4x-4x-116z-3z=60+2$$

Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.

$$-116z-3z=60+2$$

Add -116z to -3z.

$$-119z=60+2$$

Add 60 to 2.

$$-119z=62$$

Divide both sides by -119.

$$z=-\frac{62}{119}$$

Substitute -\frac{62}{119} for z in 4x+3z=-2. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+3\left(-\frac{62}{119}\right)=-2$$

Multiply 3 times -\frac{62}{119}.

$$4x-\frac{186}{119}=-2$$

Add \frac{186}{119} to both sides of the equation.

$$4x=-\frac{52}{119}$$

Divide both sides by 4.

$$x=-\frac{13}{119}$$

The system is now solved.

$$x=-\frac{13}{119},z=-\frac{62}{119}$$