Solve {l}{text{If}x/y=m/n,text{usingthecomponendoanddivide}}{qquad5x+7y/5x-7y=5m+7n/5m-7n}

$$\left. \begin{array} { l } { \text { If } \frac { x } { y } = \frac { m } { n } , \text { using the componendo and divide } } \\ { \qquad 5 x + \frac { 7 y } { 5 x } - 7 y = 5 m + \frac { 7 n } { 5 m } - 7 n } \end{array} \right.$$

Solve for x, y (complex solution)

$$x=-\frac{-25m^{2}+35mn-7n}{5\left(5m-7n\right)}$$

$$y=-\frac{n\left(-25m^{2}+35mn-7n\right)}{5m\left(5m-7n\right)}$$

$$n\neq 0\text{ and }m\neq \frac{7n}{5}\text{ and }m\neq 0\text{ and }m\neq \frac{-\sqrt{49n^{2}-28n}+7n}{10}\text{ and }m\neq \frac{\sqrt{49n^{2}-28n}+7n}{10}$$

Steps Using Substitution

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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.

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

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

$$nx+\left(-m\right)y=0$$

Add my to both sides of the equation.

$$nx=my$$

Divide both sides by n.

$$x=\frac{1}{n}my$$

Multiply \frac{1}{n} times my.

$$x=\frac{m}{n}y$$

Substitute \frac{my}{n} for x in the other equation, 25mx+\left(-35m\right)y=25m^{2}-35mn+7n.

$$25m\times \frac{m}{n}y+\left(-35m\right)y=25m^{2}-35mn+7n$$

Multiply 25m times \frac{my}{n}.

$$\frac{25m^{2}}{n}y+\left(-35m\right)y=25m^{2}-35mn+7n$$

Add \frac{25m^{2}y}{n} to -35my.

$$\left(\frac{25m^{2}}{n}-35m\right)y=25m^{2}-35mn+7n$$

Divide both sides by \frac{25m^{2}}{n}-35m.

$$y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Substitute \frac{\left(25m^{2}+7n-35nm\right)n}{5m\left(5m-7n\right)} for y in x=\frac{m}{n}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{m}{n}\times \frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Multiply \frac{m}{n} times \frac{\left(25m^{2}+7n-35nm\right)n}{5m\left(5m-7n\right)}.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)}$$

The system is now solved.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\text{, }y\neq 0$$

Steps Using Matrices

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

Write the equations in matrix form.

$$\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right).

$$inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{35m}{n\left(-35m\right)-\left(-m\right)\times 25m}&-\frac{-m}{n\left(-35m\right)-\left(-m\right)\times 25m}\\-\frac{25m}{n\left(-35m\right)-\left(-m\right)\times 25m}&\frac{n}{n\left(-35m\right)-\left(-m\right)\times 25m}\end{matrix}\right)\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{5m-7n}&\frac{1}{5\left(5m-7n\right)}\\-\frac{5}{5m-7n}&\frac{n}{5m\left(5m-7n\right)}\end{matrix}\right)\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5\left(5m-7n\right)}\left(25m^{2}-35mn+7n\right)\\\frac{n}{5m\left(5m-7n\right)}\left(25m^{2}-35mn+7n\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)}\\\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\text{, }y\neq 0$$

Steps Using Elimination

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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.

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

To make nx and 25mx equal, multiply all terms on each side of the first equation by 25m and all terms on each side of the second by n.

$$25mnx+25m\left(-m\right)y=0,n\times 25mx+n\left(-35m\right)y=n\left(25m^{2}-35mn+7n\right)$$

Simplify.

$$25mnx+\left(-25m^{2}\right)y=0,25mnx+\left(-35mn\right)y=n\left(25m^{2}-35mn+7n\right)$$

Subtract 25mnx+\left(-35mn\right)y=n\left(25m^{2}-35mn+7n\right) from 25mnx+\left(-25m^{2}\right)y=0 by subtracting like terms on each side of the equal sign.

$$25mnx+\left(-25mn\right)x+\left(-25m^{2}\right)y+35mny=-n\left(25m^{2}-35mn+7n\right)$$

Add 25mnx to -25mnx. Terms 25mnx and -25mnx cancel out, leaving an equation with only one variable that can be solved.

$$\left(-25m^{2}\right)y+35mny=-n\left(25m^{2}-35mn+7n\right)$$

Add -25m^{2}y to 35nmy.

$$5m\left(7n-5m\right)y=-n\left(25m^{2}-35mn+7n\right)$$

Divide both sides by 5m\left(-5m+7n\right).

$$y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}$$

Substitute -\frac{n\left(25m^{2}+7n-35nm\right)}{5m\left(-5m+7n\right)} for y in 25mx+\left(-35m\right)y=25m^{2}-35mn+7n. Because the resulting equation contains only one variable, you can solve for x directly.

$$25mx+\left(-35m\right)\left(-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}\right)=25m^{2}-35mn+7n$$

Multiply -35m times -\frac{n\left(25m^{2}+7n-35nm\right)}{5m\left(-5m+7n\right)}.

$$25mx+\frac{7n\left(25m^{2}-35mn+7n\right)}{7n-5m}=25m^{2}-35mn+7n$$

Subtract \frac{7n\left(25m^{2}+7n-35nm\right)}{-5m+7n} from both sides of the equation.

$$25mx=\frac{5m\left(-25m^{2}+35mn-7n\right)}{7n-5m}$$

Divide both sides by 25m.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)}$$

The system is now solved.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)},y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)},y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}\text{, }y\neq 0$$

Solve for x, y

$$x=-\frac{-25m^{2}+35mn-7n}{5\left(5m-7n\right)}$$

$$y=-\frac{n\left(-25m^{2}+35mn-7n\right)}{5m\left(5m-7n\right)}$$

$$\left(m\neq \frac{-\sqrt{49n^{2}-28n}+7n}{10}\text{ or }n>0\right)\text{ and }n\neq 0\text{ and }\left(m\neq \frac{\sqrt{49n^{2}-28n}+7n}{10}\text{ or }n>0\right)\text{ and }\left(m\neq \frac{-\sqrt{49n^{2}-28n}+7n}{10}\text{ or }n<\frac{4}{7}\right)\text{ and }\left(m\neq \frac{\sqrt{49n^{2}-28n}+7n}{10}\text{ or }n<\frac{4}{7}\right)\text{ and }m\neq 0\text{ and }m\neq \frac{7n}{5}$$

Steps Using Substitution

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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.

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

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

$$nx+\left(-m\right)y=0$$

Add my to both sides of the equation.

$$nx=my$$

Divide both sides by n.

$$x=\frac{1}{n}my$$

Multiply \frac{1}{n} times my.

$$x=\frac{m}{n}y$$

Substitute \frac{my}{n} for x in the other equation, 25mx+\left(-35m\right)y=25m^{2}-35mn+7n.

$$25m\times \frac{m}{n}y+\left(-35m\right)y=25m^{2}-35mn+7n$$

Multiply 25m times \frac{my}{n}.

$$\frac{25m^{2}}{n}y+\left(-35m\right)y=25m^{2}-35mn+7n$$

Add \frac{25m^{2}y}{n} to -35my.

$$\left(\frac{25m^{2}}{n}-35m\right)y=25m^{2}-35mn+7n$$

Divide both sides by \frac{25m^{2}}{n}-35m.

$$y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Substitute \frac{\left(25m^{2}+7n-35nm\right)n}{5m\left(5m-7n\right)} for y in x=\frac{m}{n}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{m}{n}\times \frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Multiply \frac{m}{n} times \frac{\left(25m^{2}+7n-35nm\right)n}{5m\left(5m-7n\right)}.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)}$$

The system is now solved.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\text{, }y\neq 0$$

Steps Using Matrices

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

Write the equations in matrix form.

$$\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right).

$$inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}n&-m\\25m&-35m\end{matrix}\right))\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{35m}{n\left(-35m\right)-\left(-m\right)\times 25m}&-\frac{-m}{n\left(-35m\right)-\left(-m\right)\times 25m}\\-\frac{25m}{n\left(-35m\right)-\left(-m\right)\times 25m}&\frac{n}{n\left(-35m\right)-\left(-m\right)\times 25m}\end{matrix}\right)\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{5m-7n}&\frac{1}{5\left(5m-7n\right)}\\-\frac{5}{5m-7n}&\frac{n}{5m\left(5m-7n\right)}\end{matrix}\right)\left(\begin{matrix}0\\25m^{2}-35mn+7n\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5\left(5m-7n\right)}\left(25m^{2}-35mn+7n\right)\\\frac{n}{5m\left(5m-7n\right)}\left(25m^{2}-35mn+7n\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)}\\\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{25m^{2}-35mn+7n}{5\left(5m-7n\right)},y=\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(5m-7n\right)}\text{, }y\neq 0$$

Steps Using Elimination

Consider the first equation. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ny, the least common multiple of y,n.

$$nx=ym$$

Subtract ym from both sides.

$$nx-ym=0$$

Consider the second equation. Multiply both sides of the equation by 5m.

$$5x\times 5m-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-7y\times 5m=5m\times 5m+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=5m\times 5m+7n-7n\times 5m$$

Multiply m and m to get m^{2}.

$$25xm-35ym=5m^{2}\times 5+7n-7n\times 5m$$

Multiply 5 and 5 to get 25.

$$25xm-35ym=25m^{2}+7n-7n\times 5m$$

Multiply -7 and 5 to get -35.

$$25xm-35ym=25m^{2}+7n-35nm$$

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.

$$nx+\left(-m\right)y=0,25mx+\left(-35m\right)y=25m^{2}-35mn+7n$$

To make nx and 25mx equal, multiply all terms on each side of the first equation by 25m and all terms on each side of the second by n.

$$25mnx+25m\left(-m\right)y=0,n\times 25mx+n\left(-35m\right)y=n\left(25m^{2}-35mn+7n\right)$$

Simplify.

$$25mnx+\left(-25m^{2}\right)y=0,25mnx+\left(-35mn\right)y=n\left(25m^{2}-35mn+7n\right)$$

Subtract 25mnx+\left(-35mn\right)y=n\left(25m^{2}-35mn+7n\right) from 25mnx+\left(-25m^{2}\right)y=0 by subtracting like terms on each side of the equal sign.

$$25mnx+\left(-25mn\right)x+\left(-25m^{2}\right)y+35mny=-n\left(25m^{2}-35mn+7n\right)$$

Add 25mnx to -25mnx. Terms 25mnx and -25mnx cancel out, leaving an equation with only one variable that can be solved.

$$\left(-25m^{2}\right)y+35mny=-n\left(25m^{2}-35mn+7n\right)$$

Add -25m^{2}y to 35nmy.

$$5m\left(7n-5m\right)y=-n\left(25m^{2}-35mn+7n\right)$$

Divide both sides by 5m\left(-5m+7n\right).

$$y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}$$

Substitute -\frac{n\left(25m^{2}+7n-35nm\right)}{5m\left(-5m+7n\right)} for y in 25mx+\left(-35m\right)y=25m^{2}-35mn+7n. Because the resulting equation contains only one variable, you can solve for x directly.

$$25mx+\left(-35m\right)\left(-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}\right)=25m^{2}-35mn+7n$$

Multiply -35m times -\frac{n\left(25m^{2}+7n-35nm\right)}{5m\left(-5m+7n\right)}.

$$25mx+\frac{7n\left(25m^{2}-35mn+7n\right)}{7n-5m}=25m^{2}-35mn+7n$$

Subtract \frac{7n\left(25m^{2}+7n-35nm\right)}{-5m+7n} from both sides of the equation.

$$25mx=\frac{5m\left(-25m^{2}+35mn-7n\right)}{7n-5m}$$

Divide both sides by 25m.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)}$$

The system is now solved.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)},y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}$$

Variable y cannot be equal to 0.

$$x=\frac{-25m^{2}+35mn-7n}{5\left(7n-5m\right)},y=-\frac{n\left(25m^{2}-35mn+7n\right)}{5m\left(7n-5m\right)}\text{, }y\neq 0$$