Solve {l}{x^2+y^2=1}{x+y=1.4}

$$\left\{ \begin{array} { l } { x ^ { 2 } + y ^ { 2 } = 1 } \\ { x + y = 1.4 } \end{array} \right.$$

Solve for x, y

$$x=0.8\text{, }y=0.6$$

$$x=0.6\text{, }y=0.8$$

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.

$$x+y=1.4,y^{2}+x^{2}=1$$

Solve x+y=1.4 for x by isolating x on the left hand side of the equal sign.

$$x+y=1.4$$

Subtract y from both sides of the equation.

$$x=-y+1.4$$

Substitute -y+1.4 for x in the other equation, y^{2}+x^{2}=1.

$$y^{2}+\left(-y+1.4\right)^{2}=1$$

Square -y+1.4.

$$y^{2}+y^{2}-\frac{14}{5}y+1.96=1$$

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

$$2y^{2}-\frac{14}{5}y+1.96=1$$

Subtract 1 from both sides of the equation.

$$2y^{2}-\frac{14}{5}y+0.96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\left(-1\right)^{2} for a, 1\times 1.4\left(-1\right)\times 2 for b, and 0.96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-\left(-\frac{14}{5}\right)±\sqrt{\left(-\frac{14}{5}\right)^{2}-4\times 2\times 0.96}}{2\times 2}$$

Square 1\times 1.4\left(-1\right)\times 2.

$$y=\frac{-\left(-\frac{14}{5}\right)±\sqrt{7.84-4\times 2\times 0.96}}{2\times 2}$$

Multiply -4 times 1+1\left(-1\right)^{2}.

$$y=\frac{-\left(-\frac{14}{5}\right)±\sqrt{7.84-8\times 0.96}}{2\times 2}$$

Multiply -8 times 0.96.

$$y=\frac{-\left(-\frac{14}{5}\right)±\sqrt{\frac{196-192}{25}}}{2\times 2}$$

Add 7.84 to -7.68 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{-\left(-\frac{14}{5}\right)±\sqrt{0.16}}{2\times 2}$$

Take the square root of 0.16.

$$y=\frac{-\left(-\frac{14}{5}\right)±\frac{2}{5}}{2\times 2}$$

The opposite of 1\times 1.4\left(-1\right)\times 2 is 2.8.

$$y=\frac{2.8±\frac{2}{5}}{2\times 2}$$

Multiply 2 times 1+1\left(-1\right)^{2}.

$$y=\frac{2.8±\frac{2}{5}}{4}$$

Now solve the equation y=\frac{2.8±\frac{2}{5}}{4} when ± is plus. Add 2.8 to \frac{2}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{16}{5}}{4}$$

Divide \frac{16}{5} by 4.

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

Now solve the equation y=\frac{2.8±\frac{2}{5}}{4} when ± is minus. Subtract \frac{2}{5} from 2.8 by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$y=\frac{\frac{12}{5}}{4}$$

Divide \frac{12}{5} by 4.

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

There are two solutions for y: \frac{4}{5} and \frac{3}{5}. Substitute \frac{4}{5} for y in the equation x=-y+1.4 to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{4}{5}+1.4$$

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

$$x=\frac{-4+7}{5}$$

Add -\frac{4}{5} to 1.4.

$$x=\frac{3}{5}$$

Now substitute \frac{3}{5} for y in the equation x=-y+1.4 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-\frac{3}{5}+1.4$$

Multiply -1 times \frac{3}{5}.

$$x=\frac{-3+7}{5}$$

Add -\frac{3}{5} to 1.4.

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

The system is now solved.

$$x=\frac{3}{5},y=\frac{4}{5}\text{ or }x=\frac{4}{5},y=\frac{3}{5}$$