Solve {l}{b*12x-15y=-4}{16x+10y=7}

$$\left. \begin{array} { l } { b \cdot 12 x - 15 y = - 4 } \\ { 16 x + 10 y = 7 } \end{array} \right.$$

Solve for x, y

$$x=\frac{13}{24\left(b+2\right)}$$

$$y=\frac{21b+16}{30\left(b+2\right)}$$

$$b\neq -2$$

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.

$$12bx-15y=-4,16x+10y=7$$

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

$$12bx-15y=-4$$

Add 15y to both sides of the equation.

$$12bx=15y-4$$

Divide both sides by 12b.

$$x=\frac{1}{12b}\left(15y-4\right)$$

Multiply \frac{1}{12b} times 15y-4.

$$x=\frac{5}{4b}y-\frac{1}{3b}$$

Substitute \frac{-4+15y}{12b} for x in the other equation, 16x+10y=7.

$$16\left(\frac{5}{4b}y-\frac{1}{3b}\right)+10y=7$$

Multiply 16 times \frac{-4+15y}{12b}.

$$\frac{20}{b}y-\frac{16}{3b}+10y=7$$

Add \frac{20y}{b} to 10y.

$$\left(10+\frac{20}{b}\right)y-\frac{16}{3b}=7$$

Add \frac{16}{3b} to both sides of the equation.

$$\left(10+\frac{20}{b}\right)y=7+\frac{16}{3b}$$

Divide both sides by \frac{20}{b}+10.

$$y=\frac{21b+16}{30\left(b+2\right)}$$

Substitute \frac{16+21b}{30\left(2+b\right)} for y in x=\frac{5}{4b}y-\frac{1}{3b}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{4b}\times \frac{21b+16}{30\left(b+2\right)}-\frac{1}{3b}$$

Multiply \frac{5}{4b} times \frac{16+21b}{30\left(2+b\right)}.

$$x=\frac{21b+16}{24b\left(b+2\right)}-\frac{1}{3b}$$

Add -\frac{1}{3b} to \frac{16+21b}{24b\left(2+b\right)}.

$$x=\frac{13}{24\left(b+2\right)}$$

The system is now solved.

$$x=\frac{13}{24\left(b+2\right)},y=\frac{21b+16}{30\left(b+2\right)}$$

Steps Using Matrices

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

$$12bx-15y=-4,16x+10y=7$$

Write the equations in matrix form.

$$\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\\7\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12b&-15\\16&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right))\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right))\left(\begin{matrix}-4\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right))\left(\begin{matrix}-4\\7\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12b&-15\\16&10\end{matrix}\right))\left(\begin{matrix}-4\\7\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{12b\times 10-\left(-15\times 16\right)}&-\frac{-15}{12b\times 10-\left(-15\times 16\right)}\\-\frac{16}{12b\times 10-\left(-15\times 16\right)}&\frac{12b}{12b\times 10-\left(-15\times 16\right)}\end{matrix}\right)\left(\begin{matrix}-4\\7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12\left(b+2\right)}&\frac{1}{8\left(b+2\right)}\\-\frac{2}{15\left(b+2\right)}&\frac{b}{10\left(b+2\right)}\end{matrix}\right)\left(\begin{matrix}-4\\7\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{12\left(b+2\right)}\left(-4\right)+\frac{1}{8\left(b+2\right)}\times 7\\\left(-\frac{2}{15\left(b+2\right)}\right)\left(-4\right)+\frac{b}{10\left(b+2\right)}\times 7\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{13}{24\left(b+2\right)}\\\frac{21b+16}{30\left(b+2\right)}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{13}{24\left(b+2\right)},y=\frac{21b+16}{30\left(b+2\right)}$$

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.

$$12bx-15y=-4,16x+10y=7$$

To make 12bx and 16x equal, multiply all terms on each side of the first equation by 16 and all terms on each side of the second by 12b.

$$16\times 12bx+16\left(-15\right)y=16\left(-4\right),12b\times 16x+12b\times 10y=12b\times 7$$

Simplify.

$$192bx-240y=-64,192bx+120by=84b$$

Subtract 192bx+120by=84b from 192bx-240y=-64 by subtracting like terms on each side of the equal sign.

$$192bx+\left(-192b\right)x-240y+\left(-120b\right)y=-64-84b$$

Add 192bx to -192bx. Terms 192bx and -192bx cancel out, leaving an equation with only one variable that can be solved.

$$-240y+\left(-120b\right)y=-64-84b$$

Add -240y to -120by.

$$\left(-120b-240\right)y=-64-84b$$

Add -64 to -84b.

$$\left(-120b-240\right)y=-84b-64$$

Divide both sides by -240-120b.

$$y=\frac{21b+16}{30\left(b+2\right)}$$

Substitute \frac{16+21b}{30\left(2+b\right)} for y in 16x+10y=7. Because the resulting equation contains only one variable, you can solve for x directly.

$$16x+10\times \frac{21b+16}{30\left(b+2\right)}=7$$

Multiply 10 times \frac{16+21b}{30\left(2+b\right)}.

$$16x+\frac{21b+16}{3\left(b+2\right)}=7$$

Subtract \frac{16+21b}{3\left(2+b\right)} from both sides of the equation.

$$16x=\frac{26}{3\left(b+2\right)}$$

Divide both sides by 16.

$$x=\frac{13}{24\left(b+2\right)}$$

The system is now solved.

$$x=\frac{13}{24\left(b+2\right)},y=\frac{21b+16}{30\left(b+2\right)}$$