Solve {l}{45r-2s=141}{-5r+10s=-45}

$$\left. \begin{array} { l } { 45 r - 2 s = 141 } \\ { - 5 r + 10 s = - 45 } \end{array} \right.$$

Solve for r, s

$$r=3$$

$$s=-3$$

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.

$$45r-2s=141,-5r+10s=-45$$

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

$$45r-2s=141$$

Add 2s to both sides of the equation.

$$45r=2s+141$$

Divide both sides by 45.

$$r=\frac{1}{45}\left(2s+141\right)$$

Multiply \frac{1}{45} times 2s+141.

$$r=\frac{2}{45}s+\frac{47}{15}$$

Substitute \frac{2s}{45}+\frac{47}{15} for r in the other equation, -5r+10s=-45.

$$-5\left(\frac{2}{45}s+\frac{47}{15}\right)+10s=-45$$

Multiply -5 times \frac{2s}{45}+\frac{47}{15}.

$$-\frac{2}{9}s-\frac{47}{3}+10s=-45$$

Add -\frac{2s}{9} to 10s.

$$\frac{88}{9}s-\frac{47}{3}=-45$$

Add \frac{47}{3} to both sides of the equation.

$$\frac{88}{9}s=-\frac{88}{3}$$

Divide both sides of the equation by \frac{88}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$s=-3$$

Substitute -3 for s in r=\frac{2}{45}s+\frac{47}{15}. Because the resulting equation contains only one variable, you can solve for r directly.

$$r=\frac{2}{45}\left(-3\right)+\frac{47}{15}$$

Multiply \frac{2}{45} times -3.

$$r=\frac{-2+47}{15}$$

Add \frac{47}{15} to -\frac{2}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$r=3$$

The system is now solved.

$$r=3,s=-3$$

Steps Using Matrices

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

$$45r-2s=141,-5r+10s=-45$$

Write the equations in matrix form.

$$\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right)\left(\begin{matrix}r\\s\end{matrix}\right)=\left(\begin{matrix}141\\-45\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}45&-2\\-5&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right))\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right)\left(\begin{matrix}r\\s\end{matrix}\right)=inverse(\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right))\left(\begin{matrix}141\\-45\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}r\\s\end{matrix}\right)=inverse(\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right))\left(\begin{matrix}141\\-45\end{matrix}\right)$$

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

$$\left(\begin{matrix}r\\s\end{matrix}\right)=inverse(\left(\begin{matrix}45&-2\\-5&10\end{matrix}\right))\left(\begin{matrix}141\\-45\end{matrix}\right)$$

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.

$$\left(\begin{matrix}r\\s\end{matrix}\right)=\left(\begin{matrix}\frac{10}{45\times 10-\left(-2\left(-5\right)\right)}&-\frac{-2}{45\times 10-\left(-2\left(-5\right)\right)}\\-\frac{-5}{45\times 10-\left(-2\left(-5\right)\right)}&\frac{45}{45\times 10-\left(-2\left(-5\right)\right)}\end{matrix}\right)\left(\begin{matrix}141\\-45\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}r\\s\end{matrix}\right)=\left(\begin{matrix}\frac{1}{44}&\frac{1}{220}\\\frac{1}{88}&\frac{9}{88}\end{matrix}\right)\left(\begin{matrix}141\\-45\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}r\\s\end{matrix}\right)=\left(\begin{matrix}\frac{1}{44}\times 141+\frac{1}{220}\left(-45\right)\\\frac{1}{88}\times 141+\frac{9}{88}\left(-45\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}r\\s\end{matrix}\right)=\left(\begin{matrix}3\\-3\end{matrix}\right)$$

Extract the matrix elements r and s.

$$r=3,s=-3$$

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.

$$45r-2s=141,-5r+10s=-45$$

To make 45r and -5r equal, multiply all terms on each side of the first equation by -5 and all terms on each side of the second by 45.

$$-5\times 45r-5\left(-2\right)s=-5\times 141,45\left(-5\right)r+45\times 10s=45\left(-45\right)$$

Simplify.

$$-225r+10s=-705,-225r+450s=-2025$$

Subtract -225r+450s=-2025 from -225r+10s=-705 by subtracting like terms on each side of the equal sign.

$$-225r+225r+10s-450s=-705+2025$$

Add -225r to 225r. Terms -225r and 225r cancel out, leaving an equation with only one variable that can be solved.

$$10s-450s=-705+2025$$

Add 10s to -450s.

$$-440s=-705+2025$$

Add -705 to 2025.

$$-440s=1320$$

Divide both sides by -440.

$$s=-3$$

Substitute -3 for s in -5r+10s=-45. Because the resulting equation contains only one variable, you can solve for r directly.

$$-5r+10\left(-3\right)=-45$$

Multiply 10 times -3.

$$-5r-30=-45$$

Add 30 to both sides of the equation.

$$-5r=-15$$

Divide both sides by -5.

$$r=3$$

The system is now solved.

$$r=3,s=-3$$