Solve {l}{20x_{1}-41x_{2}+21=0}{50x_{1}-20x_{2}-70=0}

$$\left\{ \begin{array} { l } { 20 x _ { 1 } - 41 x _ { 2 } + 21 = 0 } \\ { 50 x _ { 1 } - 20 x _ { 2 } - 70 = 0 } \end{array} \right.$$

Solve for x_1, x_2

$$x_{1} = \frac{329}{165} = 1\frac{164}{165} \approx 1.993939394$$

$$x_{2} = \frac{49}{33} = 1\frac{16}{33} \approx 1.484848485$$

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.

$$20x_{1}-41x_{2}+21=0,50x_{1}-20x_{2}-70=0$$

Choose one of the equations and solve it for x_{1} by isolating x_{1} on the left hand side of the equal sign.

$$20x_{1}-41x_{2}+21=0$$

Subtract 21 from both sides of the equation.

$$20x_{1}-41x_{2}=-21$$

Add 41x_{2} to both sides of the equation.

$$20x_{1}=41x_{2}-21$$

Divide both sides by 20.

$$x_{1}=\frac{1}{20}\left(41x_{2}-21\right)$$

Multiply \frac{1}{20} times 41x_{2}-21.

$$x_{1}=\frac{41}{20}x_{2}-\frac{21}{20}$$

Substitute \frac{41x_{2}-21}{20} for x_{1} in the other equation, 50x_{1}-20x_{2}-70=0.

$$50\left(\frac{41}{20}x_{2}-\frac{21}{20}\right)-20x_{2}-70=0$$

Multiply 50 times \frac{41x_{2}-21}{20}.

$$\frac{205}{2}x_{2}-\frac{105}{2}-20x_{2}-70=0$$

Add \frac{205x_{2}}{2} to -20x_{2}.

$$\frac{165}{2}x_{2}-\frac{105}{2}-70=0$$

Add -\frac{105}{2} to -70.

$$\frac{165}{2}x_{2}-\frac{245}{2}=0$$

Add \frac{245}{2} to both sides of the equation.

$$\frac{165}{2}x_{2}=\frac{245}{2}$$

Divide both sides of the equation by \frac{165}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x_{2}=\frac{49}{33}$$

Substitute \frac{49}{33} for x_{2} in x_{1}=\frac{41}{20}x_{2}-\frac{21}{20}. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$x_{1}=\frac{41}{20}\times \frac{49}{33}-\frac{21}{20}$$

Multiply \frac{41}{20} times \frac{49}{33} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x_{1}=\frac{2009}{660}-\frac{21}{20}$$

Add -\frac{21}{20} to \frac{2009}{660} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x_{1}=\frac{329}{165}$$

The system is now solved.

$$x_{1}=\frac{329}{165},x_{2}=\frac{49}{33}$$

Steps Using Matrices

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

$$20x_{1}-41x_{2}+21=0,50x_{1}-20x_{2}-70=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}-21\\70\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&-41\\50&-20\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right))\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right))\left(\begin{matrix}-21\\70\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right))\left(\begin{matrix}-21\\70\end{matrix}\right)$$

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

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=inverse(\left(\begin{matrix}20&-41\\50&-20\end{matrix}\right))\left(\begin{matrix}-21\\70\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{20}{20\left(-20\right)-\left(-41\times 50\right)}&-\frac{-41}{20\left(-20\right)-\left(-41\times 50\right)}\\-\frac{50}{20\left(-20\right)-\left(-41\times 50\right)}&\frac{20}{20\left(-20\right)-\left(-41\times 50\right)}\end{matrix}\right)\left(\begin{matrix}-21\\70\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{165}&\frac{41}{1650}\\-\frac{1}{33}&\frac{2}{165}\end{matrix}\right)\left(\begin{matrix}-21\\70\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{165}\left(-21\right)+\frac{41}{1650}\times 70\\-\frac{1}{33}\left(-21\right)+\frac{2}{165}\times 70\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x_{1}\\x_{2}\end{matrix}\right)=\left(\begin{matrix}\frac{329}{165}\\\frac{49}{33}\end{matrix}\right)$$

Extract the matrix elements x_{1} and x_{2}.

$$x_{1}=\frac{329}{165},x_{2}=\frac{49}{33}$$

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.

$$20x_{1}-41x_{2}+21=0,50x_{1}-20x_{2}-70=0$$

To make 20x_{1} and 50x_{1} equal, multiply all terms on each side of the first equation by 50 and all terms on each side of the second by 20.

$$50\times 20x_{1}+50\left(-41\right)x_{2}+50\times 21=0,20\times 50x_{1}+20\left(-20\right)x_{2}+20\left(-70\right)=0$$

Simplify.

$$1000x_{1}-2050x_{2}+1050=0,1000x_{1}-400x_{2}-1400=0$$

Subtract 1000x_{1}-400x_{2}-1400=0 from 1000x_{1}-2050x_{2}+1050=0 by subtracting like terms on each side of the equal sign.

$$1000x_{1}-1000x_{1}-2050x_{2}+400x_{2}+1050+1400=0$$

Add 1000x_{1} to -1000x_{1}. Terms 1000x_{1} and -1000x_{1} cancel out, leaving an equation with only one variable that can be solved.

$$-2050x_{2}+400x_{2}+1050+1400=0$$

Add -2050x_{2} to 400x_{2}.

$$-1650x_{2}+1050+1400=0$$

Add 1050 to 1400.

$$-1650x_{2}+2450=0$$

Subtract 2450 from both sides of the equation.

$$-1650x_{2}=-2450$$

Divide both sides by -1650.

$$x_{2}=\frac{49}{33}$$

Substitute \frac{49}{33} for x_{2} in 50x_{1}-20x_{2}-70=0. Because the resulting equation contains only one variable, you can solve for x_{1} directly.

$$50x_{1}-20\times \frac{49}{33}-70=0$$

Multiply -20 times \frac{49}{33}.

$$50x_{1}-\frac{980}{33}-70=0$$

Add -\frac{980}{33} to -70.

$$50x_{1}-\frac{3290}{33}=0$$

Add \frac{3290}{33} to both sides of the equation.

$$50x_{1}=\frac{3290}{33}$$

Divide both sides by 50.

$$x_{1}=\frac{329}{165}$$

The system is now solved.

$$x_{1}=\frac{329}{165},x_{2}=\frac{49}{33}$$