Solve {l}{x-y=4}{4x^2+7y^2=67}

$$\left. \begin{array} { l } { x - y = 4 } \\ { 4 x ^ { 2 } + 7 y ^ { 2 } = 67 } \end{array} \right.$$

Solve for x, y

$$x=\frac{45}{11}\approx 4.090909091\text{, }y=\frac{1}{11}\approx 0.090909091$$

$$x=1\text{, }y=-3$$

Steps Using Substitution and Quadratic Formula

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.

$$x-y=4,7y^{2}+4x^{2}=67$$

Solve x-y=4 for x by isolating x on the left hand side of the equal sign.

$$x-y=4$$

Subtract -y from both sides of the equation.

$$x=y+4$$

Substitute y+4 for x in the other equation, 7y^{2}+4x^{2}=67.

$$7y^{2}+4\left(y+4\right)^{2}=67$$

Square y+4.

$$7y^{2}+4\left(y^{2}+8y+16\right)=67$$

Multiply 4 times y^{2}+8y+16.

$$7y^{2}+4y^{2}+32y+64=67$$

Add 7y^{2} to 4y^{2}.

$$11y^{2}+32y+64=67$$

Subtract 67 from both sides of the equation.

$$11y^{2}+32y-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 7+4\times 1^{2} for a, 4\times 4\times 1\times 2 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-32±\sqrt{32^{2}-4\times 11\left(-3\right)}}{2\times 11}$$

Square 4\times 4\times 1\times 2.

$$y=\frac{-32±\sqrt{1024-4\times 11\left(-3\right)}}{2\times 11}$$

Multiply -4 times 7+4\times 1^{2}.

$$y=\frac{-32±\sqrt{1024-44\left(-3\right)}}{2\times 11}$$

Multiply -44 times -3.

$$y=\frac{-32±\sqrt{1024+132}}{2\times 11}$$

Add 1024 to 132.

$$y=\frac{-32±\sqrt{1156}}{2\times 11}$$

Take the square root of 1156.

$$y=\frac{-32±34}{2\times 11}$$

Multiply 2 times 7+4\times 1^{2}.

$$y=\frac{-32±34}{22}$$

Now solve the equation y=\frac{-32±34}{22} when ± is plus. Add -32 to 34.

$$y=\frac{2}{22}$$

Reduce the fraction \frac{2}{22} to lowest terms by extracting and canceling out 2.

$$y=\frac{1}{11}$$

Now solve the equation y=\frac{-32±34}{22} when ± is minus. Subtract 34 from -32.

$$y=-\frac{66}{22}$$

Divide -66 by 22.

$$y=-3$$

There are two solutions for y: \frac{1}{11} and -3. Substitute \frac{1}{11} for y in the equation x=y+4 to find the corresponding solution for x that satisfies both equations.

$$x=\frac{1}{11}+4$$

Add \frac{1}{11}\times 1 to 4.

$$x=\frac{45}{11}$$

Now substitute -3 for y in the equation x=y+4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-3+4$$

Add -3 to 4.

$$x=1$$

The system is now solved.

$$x=\frac{45}{11},y=\frac{1}{11}\text{ or }x=1,y=-3$$