Solve {l}{x/28+142-x-y/30+y/35=41/2}{frac{x}{35}+frac{142-x-y}{30}+frac{y}{28}=4frac{7}{10}}

$$\left\{ \begin{array} { l } { \frac { x } { 28 } + \frac { 142 - x - y } { 30 } + \frac { y } { 35 } = 4 \frac { 1 } { 2 } } \\ { \frac { x } { 35 } + \frac { 142 - x - y } { 30 } + \frac { y } { 28 } = 4 \frac { 7 } { 10 } } \end{array} \right.$$

Solve for x, y

$$x=42$$

$$y=70$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 420, the least common multiple of 28,30,35,2.

$$15x+14\left(142-x-y\right)+12y=210\left(4\times 2+1\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$15x+1988-14x-14y+12y=210\left(4\times 2+1\right)$$

Combine 15x and -14x to get x.

$$x+1988-14y+12y=210\left(4\times 2+1\right)$$

Combine -14y and 12y to get -2y.

$$x+1988-2y=210\left(4\times 2+1\right)$$

Multiply 4 and 2 to get 8.

$$x+1988-2y=210\left(8+1\right)$$

Add 8 and 1 to get 9.

$$x+1988-2y=210\times 9$$

Multiply 210 and 9 to get 1890.

$$x+1988-2y=1890$$

Subtract 1988 from both sides.

$$x-2y=1890-1988$$

Subtract 1988 from 1890 to get -98.

$$x-2y=-98$$

Consider the second equation. Multiply both sides of the equation by 420, the least common multiple of 35,30,28,10.

$$12x+14\left(142-x-y\right)+15y=42\left(4\times 10+7\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$12x+1988-14x-14y+15y=42\left(4\times 10+7\right)$$

Combine 12x and -14x to get -2x.

$$-2x+1988-14y+15y=42\left(4\times 10+7\right)$$

Combine -14y and 15y to get y.

$$-2x+1988+y=42\left(4\times 10+7\right)$$

Multiply 4 and 10 to get 40.

$$-2x+1988+y=42\left(40+7\right)$$

Add 40 and 7 to get 47.

$$-2x+1988+y=42\times 47$$

Multiply 42 and 47 to get 1974.

$$-2x+1988+y=1974$$

Subtract 1988 from both sides.

$$-2x+y=1974-1988$$

Subtract 1988 from 1974 to get -14.

$$-2x+y=-14$$

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.

$$x-2y=-98,-2x+y=-14$$

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

$$x-2y=-98$$

Add 2y to both sides of the equation.

$$x=2y-98$$

Substitute -98+2y for x in the other equation, -2x+y=-14.

$$-2\left(2y-98\right)+y=-14$$

Multiply -2 times -98+2y.

$$-4y+196+y=-14$$

Add -4y to y.

$$-3y+196=-14$$

Subtract 196 from both sides of the equation.

$$-3y=-210$$

Divide both sides by -3.

$$y=70$$

Substitute 70 for y in x=2y-98. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2\times 70-98$$

Multiply 2 times 70.

$$x=140-98$$

Add -98 to 140.

$$x=42$$

The system is now solved.

$$x=42,y=70$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 420, the least common multiple of 28,30,35,2.

$$15x+14\left(142-x-y\right)+12y=210\left(4\times 2+1\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$15x+1988-14x-14y+12y=210\left(4\times 2+1\right)$$

Combine 15x and -14x to get x.

$$x+1988-14y+12y=210\left(4\times 2+1\right)$$

Combine -14y and 12y to get -2y.

$$x+1988-2y=210\left(4\times 2+1\right)$$

Multiply 4 and 2 to get 8.

$$x+1988-2y=210\left(8+1\right)$$

Add 8 and 1 to get 9.

$$x+1988-2y=210\times 9$$

Multiply 210 and 9 to get 1890.

$$x+1988-2y=1890$$

Subtract 1988 from both sides.

$$x-2y=1890-1988$$

Subtract 1988 from 1890 to get -98.

$$x-2y=-98$$

Consider the second equation. Multiply both sides of the equation by 420, the least common multiple of 35,30,28,10.

$$12x+14\left(142-x-y\right)+15y=42\left(4\times 10+7\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$12x+1988-14x-14y+15y=42\left(4\times 10+7\right)$$

Combine 12x and -14x to get -2x.

$$-2x+1988-14y+15y=42\left(4\times 10+7\right)$$

Combine -14y and 15y to get y.

$$-2x+1988+y=42\left(4\times 10+7\right)$$

Multiply 4 and 10 to get 40.

$$-2x+1988+y=42\left(40+7\right)$$

Add 40 and 7 to get 47.

$$-2x+1988+y=42\times 47$$

Multiply 42 and 47 to get 1974.

$$-2x+1988+y=1974$$

Subtract 1988 from both sides.

$$-2x+y=1974-1988$$

Subtract 1988 from 1974 to get -14.

$$-2x+y=-14$$

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

$$x-2y=-98,-2x+y=-14$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right))\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right))\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right))\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-2\\-2&1\end{matrix}\right))\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-2\left(-2\right)\right)}&-\frac{-2}{1-\left(-2\left(-2\right)\right)}\\-\frac{-2}{1-\left(-2\left(-2\right)\right)}&\frac{1}{1-\left(-2\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&-\frac{2}{3}\\-\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-98\\-14\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\left(-98\right)-\frac{2}{3}\left(-14\right)\\-\frac{2}{3}\left(-98\right)-\frac{1}{3}\left(-14\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}42\\70\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=42,y=70$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 420, the least common multiple of 28,30,35,2.

$$15x+14\left(142-x-y\right)+12y=210\left(4\times 2+1\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$15x+1988-14x-14y+12y=210\left(4\times 2+1\right)$$

Combine 15x and -14x to get x.

$$x+1988-14y+12y=210\left(4\times 2+1\right)$$

Combine -14y and 12y to get -2y.

$$x+1988-2y=210\left(4\times 2+1\right)$$

Multiply 4 and 2 to get 8.

$$x+1988-2y=210\left(8+1\right)$$

Add 8 and 1 to get 9.

$$x+1988-2y=210\times 9$$

Multiply 210 and 9 to get 1890.

$$x+1988-2y=1890$$

Subtract 1988 from both sides.

$$x-2y=1890-1988$$

Subtract 1988 from 1890 to get -98.

$$x-2y=-98$$

Consider the second equation. Multiply both sides of the equation by 420, the least common multiple of 35,30,28,10.

$$12x+14\left(142-x-y\right)+15y=42\left(4\times 10+7\right)$$

Use the distributive property to multiply 14 by 142-x-y.

$$12x+1988-14x-14y+15y=42\left(4\times 10+7\right)$$

Combine 12x and -14x to get -2x.

$$-2x+1988-14y+15y=42\left(4\times 10+7\right)$$

Combine -14y and 15y to get y.

$$-2x+1988+y=42\left(4\times 10+7\right)$$

Multiply 4 and 10 to get 40.

$$-2x+1988+y=42\left(40+7\right)$$

Add 40 and 7 to get 47.

$$-2x+1988+y=42\times 47$$

Multiply 42 and 47 to get 1974.

$$-2x+1988+y=1974$$

Subtract 1988 from both sides.

$$-2x+y=1974-1988$$

Subtract 1988 from 1974 to get -14.

$$-2x+y=-14$$

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.

$$x-2y=-98,-2x+y=-14$$

To make x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 1.

$$-2x-2\left(-2\right)y=-2\left(-98\right),-2x+y=-14$$

Simplify.

$$-2x+4y=196,-2x+y=-14$$

Subtract -2x+y=-14 from -2x+4y=196 by subtracting like terms on each side of the equal sign.

$$-2x+2x+4y-y=196+14$$

Add -2x to 2x. Terms -2x and 2x cancel out, leaving an equation with only one variable that can be solved.

$$4y-y=196+14$$

Add 4y to -y.

$$3y=196+14$$

Add 196 to 14.

$$3y=210$$

Divide both sides by 3.

$$y=70$$

Substitute 70 for y in -2x+y=-14. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x+70=-14$$

Subtract 70 from both sides of the equation.

$$-2x=-84$$

Divide both sides by -2.

$$x=42$$

The system is now solved.

$$x=42,y=70$$