Solve {l}{text{If}x-y=1text{and}x^2+y^2=41,text{thenthevalueof}x+y}{{lll}{text{(a)}+9}&{text{(b)}+1}&{text{(c)}5text{or}A}}

$$\left. \begin{array} { l } { \text { If } x - y = 1 \text { and } x ^ { 2 } + y ^ { 2 } = 41 , \text { then the value of } x + y } \\ { \left. \begin{array} { l l l } { \text { (a) } + 9 } & { \text { (b) } + 1 } & { \text { (c) } 5 \text { or } A } \end{array} \right. } \end{array} \right.$$

Solve for x, y

$$x=5\text{, }y=4$$

$$x=-4\text{, }y=-5$$

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.

$$x-y=1,y^{2}+x^{2}=41$$

Solve x-y=1 for x by isolating x on the left hand side of the equal sign.

$$x-y=1$$

Subtract -y from both sides of the equation.

$$x=y+1$$

Substitute y+1 for x in the other equation, y^{2}+x^{2}=41.

$$y^{2}+\left(y+1\right)^{2}=41$$

Square y+1.

$$y^{2}+y^{2}+2y+1=41$$

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

$$2y^{2}+2y+1=41$$

Subtract 41 from both sides of the equation.

$$2y^{2}+2y-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1+1\times 1^{2} for a, 1\times 1\times 1\times 2 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$y=\frac{-2±\sqrt{2^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

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

$$y=\frac{-2±\sqrt{4-4\times 2\left(-40\right)}}{2\times 2}$$

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

$$y=\frac{-2±\sqrt{4-8\left(-40\right)}}{2\times 2}$$

Multiply -8 times -40.

$$y=\frac{-2±\sqrt{4+320}}{2\times 2}$$

Add 4 to 320.

$$y=\frac{-2±\sqrt{324}}{2\times 2}$$

Take the square root of 324.

$$y=\frac{-2±18}{2\times 2}$$

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

$$y=\frac{-2±18}{4}$$

Now solve the equation y=\frac{-2±18}{4} when ± is plus. Add -2 to 18.

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

Divide 16 by 4.

$$y=4$$

Now solve the equation y=\frac{-2±18}{4} when ± is minus. Subtract 18 from -2.

$$y=-\frac{20}{4}$$

Divide -20 by 4.

$$y=-5$$

There are two solutions for y: 4 and -5. Substitute 4 for y in the equation x=y+1 to find the corresponding solution for x that satisfies both equations.

$$x=4+1$$

Add 1\times 4 to 1.

$$x=5$$

Now substitute -5 for y in the equation x=y+1 and solve to find the corresponding solution for x that satisfies both equations.

$$x=-5+1$$

Add -5 to 1.

$$x=-4$$

The system is now solved.

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