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2\left(y+1\right)=3x-4
Consider the first equation. Variable x cannot be equal to \frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by 2\left(3x-4\right), the least common multiple of 3x-4,2.
2y+2=3x-4
Use the distributive property to multiply 2 by y+1.
2y+2-3x=-4
Subtract 3x from both sides.
2y-3x=-4-2
Subtract 2 from both sides.
2y-3x=-6
Subtract 2 from -4 to get -6.
5x+y=3x+11
Consider the second equation. Variable x cannot be equal to -\frac{11}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+11.
5x+y-3x=11
Subtract 3x from both sides.
2x+y=11
Combine 5x and -3x to get 2x.
2y-3x=-6,y+2x=11
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.
2y-3x=-6
Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.
2y=3x-6
Add 3x to both sides of the equation.
y=\frac{1}{2}\left(3x-6\right)
Divide both sides by 2.
y=\frac{3}{2}x-3
Multiply \frac{1}{2} times -6+3x.
\frac{3}{2}x-3+2x=11
Substitute \frac{3x}{2}-3 for y in the other equation, y+2x=11.
\frac{7}{2}x-3=11
Add \frac{3x}{2} to 2x.
\frac{7}{2}x=14
Add 3 to both sides of the equation.
x=4
Divide both sides of the equation by \frac{7}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.
y=\frac{3}{2}\times 4-3
Substitute 4 for x in y=\frac{3}{2}x-3. Because the resulting equation contains only one variable, you can solve for y directly.
y=6-3
Multiply \frac{3}{2} times 4.
y=3
Add -3 to 6.
y=3,x=4
The system is now solved.
2\left(y+1\right)=3x-4
Consider the first equation. Variable x cannot be equal to \frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by 2\left(3x-4\right), the least common multiple of 3x-4,2.
2y+2=3x-4
Use the distributive property to multiply 2 by y+1.
2y+2-3x=-4
Subtract 3x from both sides.
2y-3x=-4-2
Subtract 2 from both sides.
2y-3x=-6
Subtract 2 from -4 to get -6.
5x+y=3x+11
Consider the second equation. Variable x cannot be equal to -\frac{11}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+11.
5x+y-3x=11
Subtract 3x from both sides.
2x+y=11
Combine 5x and -3x to get 2x.
2y-3x=-6,y+2x=11
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-3\\1&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-6\\11\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-3\\1&2\end{matrix}\right))\left(\begin{matrix}2&-3\\1&2\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\1&2\end{matrix}\right))\left(\begin{matrix}-6\\11\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\1&2\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\1&2\end{matrix}\right))\left(\begin{matrix}-6\\11\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\1&2\end{matrix}\right))\left(\begin{matrix}-6\\11\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2\times 2-\left(-3\right)}&-\frac{-3}{2\times 2-\left(-3\right)}\\-\frac{1}{2\times 2-\left(-3\right)}&\frac{2}{2\times 2-\left(-3\right)}\end{matrix}\right)\left(\begin{matrix}-6\\11\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}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{7}&\frac{3}{7}\\-\frac{1}{7}&\frac{2}{7}\end{matrix}\right)\left(\begin{matrix}-6\\11\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{7}\left(-6\right)+\frac{3}{7}\times 11\\-\frac{1}{7}\left(-6\right)+\frac{2}{7}\times 11\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}3\\4\end{matrix}\right)
Do the arithmetic.
y=3,x=4
Extract the matrix elements y and x.
2\left(y+1\right)=3x-4
Consider the first equation. Variable x cannot be equal to \frac{4}{3} since division by zero is not defined. Multiply both sides of the equation by 2\left(3x-4\right), the least common multiple of 3x-4,2.
2y+2=3x-4
Use the distributive property to multiply 2 by y+1.
2y+2-3x=-4
Subtract 3x from both sides.
2y-3x=-4-2
Subtract 2 from both sides.
2y-3x=-6
Subtract 2 from -4 to get -6.
5x+y=3x+11
Consider the second equation. Variable x cannot be equal to -\frac{11}{3} since division by zero is not defined. Multiply both sides of the equation by 3x+11.
5x+y-3x=11
Subtract 3x from both sides.
2x+y=11
Combine 5x and -3x to get 2x.
2y-3x=-6,y+2x=11
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.
2y-3x=-6,2y+2\times 2x=2\times 11
To make 2y and y equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 2.
2y-3x=-6,2y+4x=22
Simplify.
2y-2y-3x-4x=-6-22
Subtract 2y+4x=22 from 2y-3x=-6 by subtracting like terms on each side of the equal sign.
-3x-4x=-6-22
Add 2y to -2y. Terms 2y and -2y cancel out, leaving an equation with only one variable that can be solved.
-7x=-6-22
Add -3x to -4x.
-7x=-28
Add -6 to -22.
x=4
Divide both sides by -7.
y+2\times 4=11
Substitute 4 for x in y+2x=11. Because the resulting equation contains only one variable, you can solve for y directly.
y+8=11
Multiply 2 times 4.
y=3
Subtract 8 from both sides of the equation.
y=3,x=4
The system is now solved.