Solve the following pairs of equations simultaneously 1. x+y=2 and x^2+y^2=10 2. 2x-y=6 and xy=8 3. x+y=4 and x^2+y^2=40 4. x+y=3 and x^2+2xy+y=-3 5. y=2x-1 and xy-y^2=-3 6. x-2y=1 and x^2-2xy+y^2=1 7. y=2x+1 and y^2-xy-x^2-11=0 8. x-y-3=0 and x^2-3y^2=13 9. x+y=-4 and xy=4 10. x-y=6 and (x+2y)(2x+y)=0 11. x-2y+3=0 and x^2-3y^2+2x+4y+1=0

Let's solve each pair of equations step by step:<br /><br />1. $x+y=2$ and $x^{2}+y^{2}=10$<br /> - From the first equation, we can express $y$ as $y = 2 - x$.<br /> - Substituting this into the second equation, we get $x^{2} + (2 - x)^{2} = 10$.<br /> - Simplifying, we get $x^{2} + 4 - 4x + x^{2} = 10$.<br /> - Combining like terms, we get $2x^{2} - 4x - 6 = 0$.<br /> - Solving this quadratic equation, we get $x = 3$ or $x = -1$.<br /> - Substituting these values back into $y = 2 - x$, we get $y = -1$ or $y = 3$.<br /> - Therefore, the solutions are $(x, y) = (3, -1)$ and $(x, y) = (-1, 3)$.<br /><br />2. $2x-y=6$ and $xy=8$<br /> - From the first equation, we can express $y$ as $y = 2x - 6$.<br /> - Substituting this into the second equation, we get $x(2x - 6) = 8$.<br /> - Simplifying, we get $2x^{2} - 6x - 8 = 0$.<br /> - Solving this quadratic equation, we get $x = 4$ or $x = -1$.<br /> - Substituting these values back into $y = 2x - 6$, we get $y = 2$ or $y = -8$.<br /> - Therefore, the solutions are $(x, y) = (4, 2)$ and $(x, y) = (-1, -8)$.<br /><br />3. $x+y=4$ and $x^{2}+y^{2}=40$<br /> - From the first equation, we can express $y$ as $y = 4 - x$.<br /> - Substituting this into the second equation, we get $x^{2} + (4 - x)^{2} = 40$.<br /> - Simplifying, we get $x^{2} + 16 - 8x + x^{2} = 40$.<br /> - Combining like terms, we get $2x^{2} - 8x - 24 = 0$.<br /> - Solving this quadratic equation, we get $x = 6$ or $x = -2$.<br /> - Substituting these values back into $y = 4 - x$, we get $y = -2$ or $y = 6$.<br /> - Therefore, the solutions are $(x, y) = (6, -2)$ and $(x, y) = (-2, 6)$.<br /><br />4. $x+y=3$ and $x^{2}+2xy+y=-3$<br /> - From the first equation, we can express $y$ as $y = 3 - x$.<br /> - Substituting this into the second equation, we get $x^{2} + 2x(3 - x) + (3 - x) = -3$.<br /> - Simplifying, we get $x^{2} + 6x - 2x^{2} - 6 = -3$.<br /> - Combining like terms, we get $-x^{2} + 6x - 3 = 0$.<br /> - Solving this quadratic equation, we get $x = 3$ or $x = 1$.<br /> - Substituting these values back into $y = 3 - x$, we get $y = 0$ or $y = 2$.<br /> - Therefore, the solutions are $(x, y) = (3, 0)$ and $(x, y) = (1, 2)$.<br /><br />5. $y=2x-1$ and $xy-y^{2}=-3$<br /> - Substituting $y = 2x - 1$ into the second equation, we get $x(2x - 1) - (2x - 1)^{2} = -3$.<br /> - Simplifying, we get $2x^{2} - x - 4x^{2} + 4x - 1 = -3$.<br /> - Combining like terms, we get $-2x^{2} + 3x + 1 = 0$.<br /> - Solving this quadratic equation, we get $x = 1$ or $x = -1$.<br /> -