Solve {l}{1-2y+4z-7=0}{3+5y-2z+1=0}

$$\left. \begin{array} { l } { 1 - 2 y + 4 z - 7 = 0 } \\ { 3 + 5 y - 2 z + 1 = 0 } \end{array} \right.$$

Solve for y, z

$$y=-\frac{1}{4}=-0.25$$

$$z = \frac{11}{8} = 1\frac{3}{8} = 1.375$$

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.

$$-2y+4z-6=0,5y-2z+4=0$$

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

$$-2y+4z-6=0$$

Add 6 to both sides of the equation.

$$-2y+4z=6$$

Subtract 4z from both sides of the equation.

$$-2y=-4z+6$$

Divide both sides by -2.

$$y=-\frac{1}{2}\left(-4z+6\right)$$

Multiply -\frac{1}{2} times -4z+6.

$$y=2z-3$$

Substitute 2z-3 for y in the other equation, 5y-2z+4=0.

$$5\left(2z-3\right)-2z+4=0$$

Multiply 5 times 2z-3.

$$10z-15-2z+4=0$$

Add 10z to -2z.

$$8z-15+4=0$$

Add -15 to 4.

$$8z-11=0$$

Add 11 to both sides of the equation.

$$8z=11$$

Divide both sides by 8.

$$z=\frac{11}{8}$$

Substitute \frac{11}{8} for z in y=2z-3. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=2\times \frac{11}{8}-3$$

Multiply 2 times \frac{11}{8}.

$$y=\frac{11}{4}-3$$

Add -3 to \frac{11}{4}.

$$y=-\frac{1}{4}$$

The system is now solved.

$$y=-\frac{1}{4},z=\frac{11}{8}$$

Steps Using Matrices

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

$$-2y+4z-6=0,5y-2z+4=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}6\\-4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-2&4\\5&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right))\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right))\left(\begin{matrix}6\\-4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right))\left(\begin{matrix}6\\-4\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-2&4\\5&-2\end{matrix}\right))\left(\begin{matrix}6\\-4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2\left(-2\right)-4\times 5}&-\frac{4}{-2\left(-2\right)-4\times 5}\\-\frac{5}{-2\left(-2\right)-4\times 5}&-\frac{2}{-2\left(-2\right)-4\times 5}\end{matrix}\right)\left(\begin{matrix}6\\-4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}&\frac{1}{4}\\\frac{5}{16}&\frac{1}{8}\end{matrix}\right)\left(\begin{matrix}6\\-4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8}\times 6+\frac{1}{4}\left(-4\right)\\\frac{5}{16}\times 6+\frac{1}{8}\left(-4\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{4}\\\frac{11}{8}\end{matrix}\right)$$

Extract the matrix elements y and z.

$$y=-\frac{1}{4},z=\frac{11}{8}$$

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.

$$-2y+4z-6=0,5y-2z+4=0$$

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

$$5\left(-2\right)y+5\times 4z+5\left(-6\right)=0,-2\times 5y-2\left(-2\right)z-2\times 4=0$$

Simplify.

$$-10y+20z-30=0,-10y+4z-8=0$$

Subtract -10y+4z-8=0 from -10y+20z-30=0 by subtracting like terms on each side of the equal sign.

$$-10y+10y+20z-4z-30+8=0$$

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

$$20z-4z-30+8=0$$

Add 20z to -4z.

$$16z-30+8=0$$

Add -30 to 8.

$$16z-22=0$$

Add 22 to both sides of the equation.

$$16z=22$$

Divide both sides by 16.

$$z=\frac{11}{8}$$

Substitute \frac{11}{8} for z in 5y-2z+4=0. Because the resulting equation contains only one variable, you can solve for y directly.

$$5y-2\times \frac{11}{8}+4=0$$

Multiply -2 times \frac{11}{8}.

$$5y-\frac{11}{4}+4=0$$

Add -\frac{11}{4} to 4.

$$5y+\frac{5}{4}=0$$

Subtract \frac{5}{4} from both sides of the equation.

$$5y=-\frac{5}{4}$$

Divide both sides by 5.

$$y=-\frac{1}{4}$$

The system is now solved.

$$y=-\frac{1}{4},z=\frac{11}{8}$$