Solve {l}{1/2(x/2+5/2)-(x+3/4)(x-frac{3}{4})-frac{9}{16}=-x^2-frac{y}{3}}{-frac{x}{3}-frac{y}{2}=frac{5}{2}}

$$\left\{ \begin{array} { l } { \frac { 1 } { 2 } ( \frac { x } { 2 } + \frac { 5 } { 2 } ) - ( x + \frac { 3 } { 4 } ) ( x - \frac { 3 } { 4 } ) - \frac { 9 } { 16 } = - x ^ { 2 } - \frac { y } { 3 } } \\ { - \frac { x } { 3 } - \frac { y } { 2 } = \frac { 5 } { 2 } } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=-15$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 48, the least common multiple of 2,4,16,3.

$$48\left(\frac{1}{2}\left(\frac{x}{2}+\frac{5}{2}\right)-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Since \frac{x}{2} and \frac{5}{2} have the same denominator, add them by adding their numerators.

$$48\left(\frac{1}{2}\times \frac{x+5}{2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Multiply \frac{1}{2} times \frac{x+5}{2} by multiplying numerator times numerator and denominator times denominator.

$$48\left(\frac{x+5}{2\times 2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Consider \left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{3}{4}.

$$48\left(\frac{x+5}{2\times 2}-\left(x^{2}-\frac{9}{16}\right)\right)-27=-48x^{2}-16y$$

To find the opposite of x^{2}-\frac{9}{16}, find the opposite of each term.

$$48\left(\frac{x+5}{2\times 2}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\times 2 and 16 is 16. Multiply \frac{x+5}{2\times 2} times \frac{4}{4}.

$$48\left(\frac{4\left(x+5\right)}{16}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

Since \frac{4\left(x+5\right)}{16} and \frac{9}{16} have the same denominator, add them by adding their numerators.

$$48\left(\frac{4\left(x+5\right)+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Do the multiplications in 4\left(x+5\right)+9.

$$48\left(\frac{4x+20+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Combine like terms in 4x+20+9.

$$48\left(\frac{4x+29}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Use the distributive property to multiply 48 by \frac{4x+29}{16}-x^{2}.

$$48\times \frac{4x+29}{16}-48x^{2}-27=-48x^{2}-16y$$

Cancel out 16, the greatest common factor in 48 and 16.

$$3\left(4x+29\right)-48x^{2}-27=-48x^{2}-16y$$

Use the distributive property to multiply 3 by 4x+29.

$$12x+87-48x^{2}-27=-48x^{2}-16y$$

Subtract 27 from 87 to get 60.

$$12x+60-48x^{2}=-48x^{2}-16y$$

Add 48x^{2} to both sides.

$$12x+60-48x^{2}+48x^{2}=-16y$$

Combine -48x^{2} and 48x^{2} to get 0.

$$12x+60=-16y$$

Add 16y to both sides.

$$12x+60+16y=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$12x+16y=-60$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-2x-3y=15$$

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.

$$12x+16y=-60,-2x-3y=15$$

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

$$12x+16y=-60$$

Subtract 16y from both sides of the equation.

$$12x=-16y-60$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(-16y-60\right)$$

Multiply \frac{1}{12} times -16y-60.

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

Substitute -\frac{4y}{3}-5 for x in the other equation, -2x-3y=15.

$$-2\left(-\frac{4}{3}y-5\right)-3y=15$$

Multiply -2 times -\frac{4y}{3}-5.

$$\frac{8}{3}y+10-3y=15$$

Add \frac{8y}{3} to -3y.

$$-\frac{1}{3}y+10=15$$

Subtract 10 from both sides of the equation.

$$-\frac{1}{3}y=5$$

Multiply both sides by -3.

$$y=-15$$

Substitute -15 for y in x=-\frac{4}{3}y-5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{3}\left(-15\right)-5$$

Multiply -\frac{4}{3} times -15.

$$x=20-5$$

Add -5 to 20.

$$x=15$$

The system is now solved.

$$x=15,y=-15$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 48, the least common multiple of 2,4,16,3.

$$48\left(\frac{1}{2}\left(\frac{x}{2}+\frac{5}{2}\right)-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Since \frac{x}{2} and \frac{5}{2} have the same denominator, add them by adding their numerators.

$$48\left(\frac{1}{2}\times \frac{x+5}{2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Multiply \frac{1}{2} times \frac{x+5}{2} by multiplying numerator times numerator and denominator times denominator.

$$48\left(\frac{x+5}{2\times 2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Consider \left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{3}{4}.

$$48\left(\frac{x+5}{2\times 2}-\left(x^{2}-\frac{9}{16}\right)\right)-27=-48x^{2}-16y$$

To find the opposite of x^{2}-\frac{9}{16}, find the opposite of each term.

$$48\left(\frac{x+5}{2\times 2}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\times 2 and 16 is 16. Multiply \frac{x+5}{2\times 2} times \frac{4}{4}.

$$48\left(\frac{4\left(x+5\right)}{16}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

Since \frac{4\left(x+5\right)}{16} and \frac{9}{16} have the same denominator, add them by adding their numerators.

$$48\left(\frac{4\left(x+5\right)+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Do the multiplications in 4\left(x+5\right)+9.

$$48\left(\frac{4x+20+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Combine like terms in 4x+20+9.

$$48\left(\frac{4x+29}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Use the distributive property to multiply 48 by \frac{4x+29}{16}-x^{2}.

$$48\times \frac{4x+29}{16}-48x^{2}-27=-48x^{2}-16y$$

Cancel out 16, the greatest common factor in 48 and 16.

$$3\left(4x+29\right)-48x^{2}-27=-48x^{2}-16y$$

Use the distributive property to multiply 3 by 4x+29.

$$12x+87-48x^{2}-27=-48x^{2}-16y$$

Subtract 27 from 87 to get 60.

$$12x+60-48x^{2}=-48x^{2}-16y$$

Add 48x^{2} to both sides.

$$12x+60-48x^{2}+48x^{2}=-16y$$

Combine -48x^{2} and 48x^{2} to get 0.

$$12x+60=-16y$$

Add 16y to both sides.

$$12x+60+16y=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$12x+16y=-60$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-2x-3y=15$$

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

$$12x+16y=-60,-2x-3y=15$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-60\\15\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&16\\-2&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right))\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right))\left(\begin{matrix}-60\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right))\left(\begin{matrix}-60\\15\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&16\\-2&-3\end{matrix}\right))\left(\begin{matrix}-60\\15\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{12\left(-3\right)-16\left(-2\right)}&-\frac{16}{12\left(-3\right)-16\left(-2\right)}\\-\frac{-2}{12\left(-3\right)-16\left(-2\right)}&\frac{12}{12\left(-3\right)-16\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}-60\\15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}&4\\-\frac{1}{2}&-3\end{matrix}\right)\left(\begin{matrix}-60\\15\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{4}\left(-60\right)+4\times 15\\-\frac{1}{2}\left(-60\right)-3\times 15\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\-15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=-15$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 48, the least common multiple of 2,4,16,3.

$$48\left(\frac{1}{2}\left(\frac{x}{2}+\frac{5}{2}\right)-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Since \frac{x}{2} and \frac{5}{2} have the same denominator, add them by adding their numerators.

$$48\left(\frac{1}{2}\times \frac{x+5}{2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Multiply \frac{1}{2} times \frac{x+5}{2} by multiplying numerator times numerator and denominator times denominator.

$$48\left(\frac{x+5}{2\times 2}-\left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right)\right)-27=-48x^{2}-16y$$

Consider \left(x+\frac{3}{4}\right)\left(x-\frac{3}{4}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square \frac{3}{4}.

$$48\left(\frac{x+5}{2\times 2}-\left(x^{2}-\frac{9}{16}\right)\right)-27=-48x^{2}-16y$$

To find the opposite of x^{2}-\frac{9}{16}, find the opposite of each term.

$$48\left(\frac{x+5}{2\times 2}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2\times 2 and 16 is 16. Multiply \frac{x+5}{2\times 2} times \frac{4}{4}.

$$48\left(\frac{4\left(x+5\right)}{16}-x^{2}+\frac{9}{16}\right)-27=-48x^{2}-16y$$

Since \frac{4\left(x+5\right)}{16} and \frac{9}{16} have the same denominator, add them by adding their numerators.

$$48\left(\frac{4\left(x+5\right)+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Do the multiplications in 4\left(x+5\right)+9.

$$48\left(\frac{4x+20+9}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Combine like terms in 4x+20+9.

$$48\left(\frac{4x+29}{16}-x^{2}\right)-27=-48x^{2}-16y$$

Use the distributive property to multiply 48 by \frac{4x+29}{16}-x^{2}.

$$48\times \frac{4x+29}{16}-48x^{2}-27=-48x^{2}-16y$$

Cancel out 16, the greatest common factor in 48 and 16.

$$3\left(4x+29\right)-48x^{2}-27=-48x^{2}-16y$$

Use the distributive property to multiply 3 by 4x+29.

$$12x+87-48x^{2}-27=-48x^{2}-16y$$

Subtract 27 from 87 to get 60.

$$12x+60-48x^{2}=-48x^{2}-16y$$

Add 48x^{2} to both sides.

$$12x+60-48x^{2}+48x^{2}=-16y$$

Combine -48x^{2} and 48x^{2} to get 0.

$$12x+60=-16y$$

Add 16y to both sides.

$$12x+60+16y=0$$

Subtract 60 from both sides. Anything subtracted from zero gives its negation.

$$12x+16y=-60$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$-2x-3y=15$$

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.

$$12x+16y=-60,-2x-3y=15$$

To make 12x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 12.

$$-2\times 12x-2\times 16y=-2\left(-60\right),12\left(-2\right)x+12\left(-3\right)y=12\times 15$$

Simplify.

$$-24x-32y=120,-24x-36y=180$$

Subtract -24x-36y=180 from -24x-32y=120 by subtracting like terms on each side of the equal sign.

$$-24x+24x-32y+36y=120-180$$

Add -24x to 24x. Terms -24x and 24x cancel out, leaving an equation with only one variable that can be solved.

$$-32y+36y=120-180$$

Add -32y to 36y.

$$4y=120-180$$

Add 120 to -180.

$$4y=-60$$

Divide both sides by 4.

$$y=-15$$

Substitute -15 for y in -2x-3y=15. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x-3\left(-15\right)=15$$

Multiply -3 times -15.

$$-2x+45=15$$

Subtract 45 from both sides of the equation.

$$-2x=-30$$

Divide both sides by -2.

$$x=15$$

The system is now solved.

$$x=15,y=-15$$