Solve {l}{6-5y+2z/3=16+2y-3z/2}{6-5y+2z/3=1-5y+4z/4}

$$\left. \begin{array} { l } { \frac { 6 - 5 y + 2 z } { 3 } = \frac { 16 + 2 y - 3 z } { 2 } } \\ { \frac { 6 - 5 y + 2 z } { 3 } = \frac { 1 - 5 y + 4 z } { 4 } } \end{array} \right.$$

Solve for y, z

$$y=1$$

$$z=4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\left(6-5y+2z\right)=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 2 by 6-5y+2z.

$$12-10y+4z=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 3 by 16+2y-3z.

$$12-10y+4z=48+6y-9z$$

Subtract 6y from both sides.

$$12-10y+4z-6y=48-9z$$

Combine -10y and -6y to get -16y.

$$12-16y+4z=48-9z$$

Add 9z to both sides.

$$12-16y+4z+9z=48$$

Combine 4z and 9z to get 13z.

$$12-16y+13z=48$$

Subtract 12 from both sides.

$$-16y+13z=48-12$$

Subtract 12 from 48 to get 36.

$$-16y+13z=36$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\left(6-5y+2z\right)=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 4 by 6-5y+2z.

$$24-20y+8z=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 3 by 1-5y+4z.

$$24-20y+8z=3-15y+12z$$

Add 15y to both sides.

$$24-20y+8z+15y=3+12z$$

Combine -20y and 15y to get -5y.

$$24-5y+8z=3+12z$$

Subtract 12z from both sides.

$$24-5y+8z-12z=3$$

Combine 8z and -12z to get -4z.

$$24-5y-4z=3$$

Subtract 24 from both sides.

$$-5y-4z=3-24$$

Subtract 24 from 3 to get -21.

$$-5y-4z=-21$$

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.

$$-16y+13z=36,-5y-4z=-21$$

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

$$-16y+13z=36$$

Subtract 13z from both sides of the equation.

$$-16y=-13z+36$$

Divide both sides by -16.

$$y=-\frac{1}{16}\left(-13z+36\right)$$

Multiply -\frac{1}{16} times -13z+36.

$$y=\frac{13}{16}z-\frac{9}{4}$$

Substitute \frac{13z}{16}-\frac{9}{4} for y in the other equation, -5y-4z=-21.

$$-5\left(\frac{13}{16}z-\frac{9}{4}\right)-4z=-21$$

Multiply -5 times \frac{13z}{16}-\frac{9}{4}.

$$-\frac{65}{16}z+\frac{45}{4}-4z=-21$$

Add -\frac{65z}{16} to -4z.

$$-\frac{129}{16}z+\frac{45}{4}=-21$$

Subtract \frac{45}{4} from both sides of the equation.

$$-\frac{129}{16}z=-\frac{129}{4}$$

Divide both sides of the equation by -\frac{129}{16}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$z=4$$

Substitute 4 for z in y=\frac{13}{16}z-\frac{9}{4}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{13}{16}\times 4-\frac{9}{4}$$

Multiply \frac{13}{16} times 4.

$$y=\frac{13-9}{4}$$

Add -\frac{9}{4} to \frac{13}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=1$$

The system is now solved.

$$y=1,z=4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\left(6-5y+2z\right)=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 2 by 6-5y+2z.

$$12-10y+4z=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 3 by 16+2y-3z.

$$12-10y+4z=48+6y-9z$$

Subtract 6y from both sides.

$$12-10y+4z-6y=48-9z$$

Combine -10y and -6y to get -16y.

$$12-16y+4z=48-9z$$

Add 9z to both sides.

$$12-16y+4z+9z=48$$

Combine 4z and 9z to get 13z.

$$12-16y+13z=48$$

Subtract 12 from both sides.

$$-16y+13z=48-12$$

Subtract 12 from 48 to get 36.

$$-16y+13z=36$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\left(6-5y+2z\right)=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 4 by 6-5y+2z.

$$24-20y+8z=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 3 by 1-5y+4z.

$$24-20y+8z=3-15y+12z$$

Add 15y to both sides.

$$24-20y+8z+15y=3+12z$$

Combine -20y and 15y to get -5y.

$$24-5y+8z=3+12z$$

Subtract 12z from both sides.

$$24-5y+8z-12z=3$$

Combine 8z and -12z to get -4z.

$$24-5y-4z=3$$

Subtract 24 from both sides.

$$-5y-4z=3-24$$

Subtract 24 from 3 to get -21.

$$-5y-4z=-21$$

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

$$-16y+13z=36,-5y-4z=-21$$

Write the equations in matrix form.

$$\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}36\\-21\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right))\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right))\left(\begin{matrix}36\\-21\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right))\left(\begin{matrix}36\\-21\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\z\end{matrix}\right)=inverse(\left(\begin{matrix}-16&13\\-5&-4\end{matrix}\right))\left(\begin{matrix}36\\-21\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{-16\left(-4\right)-13\left(-5\right)}&-\frac{13}{-16\left(-4\right)-13\left(-5\right)}\\-\frac{-5}{-16\left(-4\right)-13\left(-5\right)}&-\frac{16}{-16\left(-4\right)-13\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}36\\-21\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{129}&-\frac{13}{129}\\\frac{5}{129}&-\frac{16}{129}\end{matrix}\right)\left(\begin{matrix}36\\-21\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{129}\times 36-\frac{13}{129}\left(-21\right)\\\frac{5}{129}\times 36-\frac{16}{129}\left(-21\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements y and z.

$$y=1,z=4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2\left(6-5y+2z\right)=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 2 by 6-5y+2z.

$$12-10y+4z=3\left(16+2y-3z\right)$$

Use the distributive property to multiply 3 by 16+2y-3z.

$$12-10y+4z=48+6y-9z$$

Subtract 6y from both sides.

$$12-10y+4z-6y=48-9z$$

Combine -10y and -6y to get -16y.

$$12-16y+4z=48-9z$$

Add 9z to both sides.

$$12-16y+4z+9z=48$$

Combine 4z and 9z to get 13z.

$$12-16y+13z=48$$

Subtract 12 from both sides.

$$-16y+13z=48-12$$

Subtract 12 from 48 to get 36.

$$-16y+13z=36$$

Consider the second equation. Multiply both sides of the equation by 12, the least common multiple of 3,4.

$$4\left(6-5y+2z\right)=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 4 by 6-5y+2z.

$$24-20y+8z=3\left(1-5y+4z\right)$$

Use the distributive property to multiply 3 by 1-5y+4z.

$$24-20y+8z=3-15y+12z$$

Add 15y to both sides.

$$24-20y+8z+15y=3+12z$$

Combine -20y and 15y to get -5y.

$$24-5y+8z=3+12z$$

Subtract 12z from both sides.

$$24-5y+8z-12z=3$$

Combine 8z and -12z to get -4z.

$$24-5y-4z=3$$

Subtract 24 from both sides.

$$-5y-4z=3-24$$

Subtract 24 from 3 to get -21.

$$-5y-4z=-21$$

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.

$$-16y+13z=36,-5y-4z=-21$$

To make -16y 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 -16.

$$-5\left(-16\right)y-5\times 13z=-5\times 36,-16\left(-5\right)y-16\left(-4\right)z=-16\left(-21\right)$$

Simplify.

$$80y-65z=-180,80y+64z=336$$

Subtract 80y+64z=336 from 80y-65z=-180 by subtracting like terms on each side of the equal sign.

$$80y-80y-65z-64z=-180-336$$

Add 80y to -80y. Terms 80y and -80y cancel out, leaving an equation with only one variable that can be solved.

$$-65z-64z=-180-336$$

Add -65z to -64z.

$$-129z=-180-336$$

Add -180 to -336.

$$-129z=-516$$

Divide both sides by -129.

$$z=4$$

Substitute 4 for z in -5y-4z=-21. Because the resulting equation contains only one variable, you can solve for y directly.

$$-5y-4\times 4=-21$$

Multiply -4 times 4.

$$-5y-16=-21$$

Add 16 to both sides of the equation.

$$-5y=-5$$

Divide both sides by -5.

$$y=1$$

The system is now solved.

$$y=1,z=4$$