Solve درج ذیل ہمزاد مساواتوں کو حل کیجئے x + y = 6; 2x ^ 2 - 4y ^ 2 = 50 ) | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$2 x ^ { 2 } - 4 y ^ { 2 } = 50 \quad , \quad x + y = 6$$

Solve for x, y

$x=\sqrt{47}+12\approx 18.8556546\text{, }y=-\left(\sqrt{47}+6\right)\approx -12.8556546$
$x=12-\sqrt{47}\approx 5.1443454\text{, }y=\sqrt{47}-6\approx 0.8556546$

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.

$$x+y=6,-4y^{2}+2x^{2}=50$$

Solve $x+y=6$ for $x$ by isolating $x$ on the left hand side of the equal sign.

$$x+y=6$$

Subtract $y$ from both sides of the equation.

$$x=-y+6$$

Substitute $-y+6$ for $x$ in the other equation, $-4y^{2}+2x^{2}=50$.

$$-4y^{2}+2\left(-y+6\right)^{2}=50$$

Square $-y+6$.

$$-4y^{2}+2\left(y^{2}-12y+36\right)=50$$

Multiply $2$ times $y^{2}-12y+36$.

$$-4y^{2}+2y^{2}-24y+72=50$$

Add $-4y^{2}$ to $2y^{2}$.

$$-2y^{2}-24y+72=50$$

Subtract $50$ from both sides of the equation.

$$-2y^{2}-24y+22=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-4+2\left(-1\right)^{2}$ for $a$, $2\times 6\left(-1\right)\times 2$ for $b$, and $22$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\left(-2\right)\times 22}}{2\left(-2\right)}$$

Square $2\times 6\left(-1\right)\times 2$.

$$y=\frac{-\left(-24\right)±\sqrt{576-4\left(-2\right)\times 22}}{2\left(-2\right)}$$

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

$$y=\frac{-\left(-24\right)±\sqrt{576+8\times 22}}{2\left(-2\right)}$$

Multiply $8$ times $22$.

$$y=\frac{-\left(-24\right)±\sqrt{576+176}}{2\left(-2\right)}$$

Add $576$ to $176$.

$$y=\frac{-\left(-24\right)±\sqrt{752}}{2\left(-2\right)}$$

Take the square root of $752$.

$$y=\frac{-\left(-24\right)±4\sqrt{47}}{2\left(-2\right)}$$

The opposite of $2\times 6\left(-1\right)\times 2$ is $24$.

$$y=\frac{24±4\sqrt{47}}{2\left(-2\right)}$$

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

$$y=\frac{24±4\sqrt{47}}{-4}$$

Now solve the equation $y=\frac{24±4\sqrt{47}}{-4}$ when $±$ is plus. Add $24$ to $4\sqrt{47}$.

$$y=\frac{4\sqrt{47}+24}{-4}$$

Divide $24+4\sqrt{47}$ by $-4$.

$$y=-\left(\sqrt{47}+6\right)$$

Now solve the equation $y=\frac{24±4\sqrt{47}}{-4}$ when $±$ is minus. Subtract $4\sqrt{47}$ from $24$.

$$y=\frac{24-4\sqrt{47}}{-4}$$

Divide $24-4\sqrt{47}$ by $-4$.

$$y=\sqrt{47}-6$$

There are two solutions for $y$: $-\left(6+\sqrt{47}\right)$ and $-6+\sqrt{47}$. Substitute $-\left(6+\sqrt{47}\right)$ for $y$ in the equation $x=-y+6$ to find the corresponding solution for $x$ that satisfies both equations.

$$x=-\left(-\left(\sqrt{47}+6\right)\right)+6$$

Now substitute $-6+\sqrt{47}$ for $y$ in the equation $x=-y+6$ and solve to find the corresponding solution for $x$ that satisfies both equations.

$$x=-\left(\sqrt{47}-6\right)+6$$

The system is now solved.

$$x=-\left(-\left(\sqrt{47}+6\right)\right)+6,y=-\left(\sqrt{47}+6\right)\text{ or }x=-\left(\sqrt{47}-6\right)+6,y=\sqrt{47}-6$$