Solve {l}{(x+2)^2+1=x^2+5y}{3x+y=1}

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

Solve for x, y

$$x=0$$

$$y=1$$

Steps Using Substitution

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+4x+4+1=x^{2}+5y$$

Add 4 and 1 to get 5.

$$x^{2}+4x+5=x^{2}+5y$$

Subtract x^{2} from both sides.

$$x^{2}+4x+5-x^{2}=5y$$

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

$$4x+5=5y$$

Subtract 5y from both sides.

$$4x+5-5y=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$4x-5y=-5$$

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.

$$4x-5y=-5,3x+y=1$$

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

$$4x-5y=-5$$

Add 5y to both sides of the equation.

$$4x=5y-5$$

Divide both sides by 4.

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

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

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

Substitute \frac{-5+5y}{4} for x in the other equation, 3x+y=1.

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

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

$$\frac{15}{4}y-\frac{15}{4}+y=1$$

Add \frac{15y}{4} to y.

$$\frac{19}{4}y-\frac{15}{4}=1$$

Add \frac{15}{4} to both sides of the equation.

$$\frac{19}{4}y=\frac{19}{4}$$

Divide both sides of the equation by \frac{19}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=1$$

Substitute 1 for y in x=\frac{5}{4}y-\frac{5}{4}. Because the resulting equation contains only one variable, you can solve for x directly.

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

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

$$x=0$$

The system is now solved.

$$x=0,y=1$$

Steps Using Matrices

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+4x+4+1=x^{2}+5y$$

Add 4 and 1 to get 5.

$$x^{2}+4x+5=x^{2}+5y$$

Subtract x^{2} from both sides.

$$x^{2}+4x+5-x^{2}=5y$$

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

$$4x+5=5y$$

Subtract 5y from both sides.

$$4x+5-5y=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$4x-5y=-5$$

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

$$4x-5y=-5,3x+y=1$$

Write the equations in matrix form.

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

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-5\\3&1\end{matrix}\right).

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

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

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

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

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

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4-\left(-5\times 3\right)}&-\frac{-5}{4-\left(-5\times 3\right)}\\-\frac{3}{4-\left(-5\times 3\right)}&\frac{4}{4-\left(-5\times 3\right)}\end{matrix}\right)\left(\begin{matrix}-5\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{19}&\frac{5}{19}\\-\frac{3}{19}&\frac{4}{19}\end{matrix}\right)\left(\begin{matrix}-5\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{19}\left(-5\right)+\frac{5}{19}\\-\frac{3}{19}\left(-5\right)+\frac{4}{19}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=1$$

Steps Using Elimination

Consider the first equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+2\right)^{2}.

$$x^{2}+4x+4+1=x^{2}+5y$$

Add 4 and 1 to get 5.

$$x^{2}+4x+5=x^{2}+5y$$

Subtract x^{2} from both sides.

$$x^{2}+4x+5-x^{2}=5y$$

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

$$4x+5=5y$$

Subtract 5y from both sides.

$$4x+5-5y=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$4x-5y=-5$$

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.

$$4x-5y=-5,3x+y=1$$

To make 4x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 4.

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

Simplify.

$$12x-15y=-15,12x+4y=4$$

Subtract 12x+4y=4 from 12x-15y=-15 by subtracting like terms on each side of the equal sign.

$$12x-12x-15y-4y=-15-4$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$-15y-4y=-15-4$$

Add -15y to -4y.

$$-19y=-15-4$$

Add -15 to -4.

$$-19y=-19$$

Divide both sides by -19.

$$y=1$$

Substitute 1 for y in 3x+y=1. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x+1=1$$

Subtract 1 from both sides of the equation.

$$3x=0$$

Divide both sides by 3.

$$x=0$$

The system is now solved.

$$x=0,y=1$$