Solve {l}{x-7(y-1)=-2(5/2+x)}{-2(1-2x-y)=1-x-5}

$$\left\{ \begin{array} { l } { x - 7 ( y - 1 ) = - 2 ( \frac { 5 } { 2 } + x ) } \\ { - 2 ( 1 - 2 x - y ) = 1 - x - 5 } \end{array} \right.$$

Solve for x, y

$$x=-\frac{38}{41}\approx -0.926829268$$

$$y = \frac{54}{41} = 1\frac{13}{41} \approx 1.317073171$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply -7 by y-1.

$$x-7y+7=-2\left(\frac{5}{2}+x\right)$$

Use the distributive property to multiply -2 by \frac{5}{2}+x.

$$x-7y+7=-5-2x$$

Add 2x to both sides.

$$x-7y+7+2x=-5$$

Combine x and 2x to get 3x.

$$3x-7y+7=-5$$

Subtract 7 from both sides.

$$3x-7y=-5-7$$

Subtract 7 from -5 to get -12.

$$3x-7y=-12$$

Consider the second equation. Use the distributive property to multiply -2 by 1-2x-y.

$$-2+4x+2y=1-x-5$$

Subtract 5 from 1 to get -4.

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

Add x to both sides.

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

Combine 4x and x to get 5x.

$$-2+5x+2y=-4$$

Add 2 to both sides.

$$5x+2y=-4+2$$

Add -4 and 2 to get -2.

$$5x+2y=-2$$

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.

$$3x-7y=-12,5x+2y=-2$$

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

$$3x-7y=-12$$

Add 7y to both sides of the equation.

$$3x=7y-12$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(7y-12\right)$$

Multiply \frac{1}{3} times 7y-12.

$$x=\frac{7}{3}y-4$$

Substitute \frac{7y}{3}-4 for x in the other equation, 5x+2y=-2.

$$5\left(\frac{7}{3}y-4\right)+2y=-2$$

Multiply 5 times \frac{7y}{3}-4.

$$\frac{35}{3}y-20+2y=-2$$

Add \frac{35y}{3} to 2y.

$$\frac{41}{3}y-20=-2$$

Add 20 to both sides of the equation.

$$\frac{41}{3}y=18$$

Divide both sides of the equation by \frac{41}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{54}{41}$$

Substitute \frac{54}{41} for y in x=\frac{7}{3}y-4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{7}{3}\times \frac{54}{41}-4$$

Multiply \frac{7}{3} times \frac{54}{41} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{126}{41}-4$$

Add -4 to \frac{126}{41}.

$$x=-\frac{38}{41}$$

The system is now solved.

$$x=-\frac{38}{41},y=\frac{54}{41}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply -7 by y-1.

$$x-7y+7=-2\left(\frac{5}{2}+x\right)$$

Use the distributive property to multiply -2 by \frac{5}{2}+x.

$$x-7y+7=-5-2x$$

Add 2x to both sides.

$$x-7y+7+2x=-5$$

Combine x and 2x to get 3x.

$$3x-7y+7=-5$$

Subtract 7 from both sides.

$$3x-7y=-5-7$$

Subtract 7 from -5 to get -12.

$$3x-7y=-12$$

Consider the second equation. Use the distributive property to multiply -2 by 1-2x-y.

$$-2+4x+2y=1-x-5$$

Subtract 5 from 1 to get -4.

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

Add x to both sides.

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

Combine 4x and x to get 5x.

$$-2+5x+2y=-4$$

Add 2 to both sides.

$$5x+2y=-4+2$$

Add -4 and 2 to get -2.

$$5x+2y=-2$$

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

$$3x-7y=-12,5x+2y=-2$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-7\\5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-7\\5&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-7\\5&2\end{matrix}\right))\left(\begin{matrix}3&-7\\5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-7\\5&2\end{matrix}\right))\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-7\\5&2\end{matrix}\right))\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-7\\5&2\end{matrix}\right))\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3\times 2-\left(-7\times 5\right)}&-\frac{-7}{3\times 2-\left(-7\times 5\right)}\\-\frac{5}{3\times 2-\left(-7\times 5\right)}&\frac{3}{3\times 2-\left(-7\times 5\right)}\end{matrix}\right)\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{41}&\frac{7}{41}\\-\frac{5}{41}&\frac{3}{41}\end{matrix}\right)\left(\begin{matrix}-12\\-2\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{41}\left(-12\right)+\frac{7}{41}\left(-2\right)\\-\frac{5}{41}\left(-12\right)+\frac{3}{41}\left(-2\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{38}{41}\\\frac{54}{41}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{38}{41},y=\frac{54}{41}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply -7 by y-1.

$$x-7y+7=-2\left(\frac{5}{2}+x\right)$$

Use the distributive property to multiply -2 by \frac{5}{2}+x.

$$x-7y+7=-5-2x$$

Add 2x to both sides.

$$x-7y+7+2x=-5$$

Combine x and 2x to get 3x.

$$3x-7y+7=-5$$

Subtract 7 from both sides.

$$3x-7y=-5-7$$

Subtract 7 from -5 to get -12.

$$3x-7y=-12$$

Consider the second equation. Use the distributive property to multiply -2 by 1-2x-y.

$$-2+4x+2y=1-x-5$$

Subtract 5 from 1 to get -4.

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

Add x to both sides.

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

Combine 4x and x to get 5x.

$$-2+5x+2y=-4$$

Add 2 to both sides.

$$5x+2y=-4+2$$

Add -4 and 2 to get -2.

$$5x+2y=-2$$

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.

$$3x-7y=-12,5x+2y=-2$$

To make 3x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 3.

$$5\times 3x+5\left(-7\right)y=5\left(-12\right),3\times 5x+3\times 2y=3\left(-2\right)$$

Simplify.

$$15x-35y=-60,15x+6y=-6$$

Subtract 15x+6y=-6 from 15x-35y=-60 by subtracting like terms on each side of the equal sign.

$$15x-15x-35y-6y=-60+6$$

Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.

$$-35y-6y=-60+6$$

Add -35y to -6y.

$$-41y=-60+6$$

Add -60 to 6.

$$-41y=-54$$

Divide both sides by -41.

$$y=\frac{54}{41}$$

Substitute \frac{54}{41} for y in 5x+2y=-2. Because the resulting equation contains only one variable, you can solve for x directly.

$$5x+2\times \frac{54}{41}=-2$$

Multiply 2 times \frac{54}{41}.

$$5x+\frac{108}{41}=-2$$

Subtract \frac{108}{41} from both sides of the equation.

$$5x=-\frac{190}{41}$$

Divide both sides by 5.

$$x=-\frac{38}{41}$$

The system is now solved.

$$x=-\frac{38}{41},y=\frac{54}{41}$$