Solve {l}{4x-3y=5}{100x+y=5314y}

$$\left. \begin{array} { l } { 4 x - 3 y = 5 } \\ { 100 x + y = 5314 y } \end{array} \right.$$

Solve for x, y

$$x = \frac{8855}{6984} = 1\frac{1871}{6984} \approx 1.267898053$$

$$y=\frac{125}{5238}\approx 0.02386407$$

Steps Using Substitution

Consider the second equation. Subtract 5314y from both sides.

$$100x+y-5314y=0$$

Combine y and -5314y to get -5313y.

$$100x-5313y=0$$

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.

$$4x-3y=5,100x-5313y=0$$

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

$$4x-3y=5$$

Add 3y to both sides of the equation.

$$4x=3y+5$$

Divide both sides by 4.

$$x=\frac{1}{4}\left(3y+5\right)$$

Multiply \frac{1}{4} times 3y+5.

$$x=\frac{3}{4}y+\frac{5}{4}$$

Substitute \frac{3y+5}{4} for x in the other equation, 100x-5313y=0.

$$100\left(\frac{3}{4}y+\frac{5}{4}\right)-5313y=0$$

Multiply 100 times \frac{3y+5}{4}.

$$75y+125-5313y=0$$

Add 75y to -5313y.

$$-5238y+125=0$$

Subtract 125 from both sides of the equation.

$$-5238y=-125$$

Divide both sides by -5238.

$$y=\frac{125}{5238}$$

Substitute \frac{125}{5238} for y in x=\frac{3}{4}y+\frac{5}{4}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{4}\times \frac{125}{5238}+\frac{5}{4}$$

Multiply \frac{3}{4} times \frac{125}{5238} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{125}{6984}+\frac{5}{4}$$

Add \frac{5}{4} to \frac{125}{6984} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{8855}{6984}$$

The system is now solved.

$$x=\frac{8855}{6984},y=\frac{125}{5238}$$

Steps Using Matrices

Consider the second equation. Subtract 5314y from both sides.

$$100x+y-5314y=0$$

Combine y and -5314y to get -5313y.

$$100x-5313y=0$$

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

$$4x-3y=5,100x-5313y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right))\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right))\left(\begin{matrix}5\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right))\left(\begin{matrix}5\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}4&-3\\100&-5313\end{matrix}\right))\left(\begin{matrix}5\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5313}{4\left(-5313\right)-\left(-3\times 100\right)}&-\frac{-3}{4\left(-5313\right)-\left(-3\times 100\right)}\\-\frac{100}{4\left(-5313\right)-\left(-3\times 100\right)}&\frac{4}{4\left(-5313\right)-\left(-3\times 100\right)}\end{matrix}\right)\left(\begin{matrix}5\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1771}{6984}&-\frac{1}{6984}\\\frac{25}{5238}&-\frac{1}{5238}\end{matrix}\right)\left(\begin{matrix}5\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1771}{6984}\times 5\\\frac{25}{5238}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8855}{6984}\\\frac{125}{5238}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{8855}{6984},y=\frac{125}{5238}$$

Steps Using Elimination

Consider the second equation. Subtract 5314y from both sides.

$$100x+y-5314y=0$$

Combine y and -5314y to get -5313y.

$$100x-5313y=0$$

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.

$$4x-3y=5,100x-5313y=0$$

To make 4x and 100x equal, multiply all terms on each side of the first equation by 100 and all terms on each side of the second by 4.

$$100\times 4x+100\left(-3\right)y=100\times 5,4\times 100x+4\left(-5313\right)y=0$$

Simplify.

$$400x-300y=500,400x-21252y=0$$

Subtract 400x-21252y=0 from 400x-300y=500 by subtracting like terms on each side of the equal sign.

$$400x-400x-300y+21252y=500$$

Add 400x to -400x. Terms 400x and -400x cancel out, leaving an equation with only one variable that can be solved.

$$-300y+21252y=500$$

Add -300y to 21252y.

$$20952y=500$$

Divide both sides by 20952.

$$y=\frac{125}{5238}$$

Substitute \frac{125}{5238} for y in 100x-5313y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$100x-5313\times \frac{125}{5238}=0$$

Multiply -5313 times \frac{125}{5238}.

$$100x-\frac{221375}{1746}=0$$

Add \frac{221375}{1746} to both sides of the equation.

$$100x=\frac{221375}{1746}$$

Divide both sides by 100.

$$x=\frac{8855}{6984}$$

The system is now solved.

$$x=\frac{8855}{6984},y=\frac{125}{5238}$$