Question

√6+ x - 3 √3y + 4 = -10, 4√3y+ 4 - 5 √6 + x = 6;​

Answer 1

$\displaystyle\bf\\\left \{ {{\sqrt{6+x} -3\sqrt{3y+4}=-10} \atop {4\sqrt{3y+4} -5\sqrt{6+x} =6}} \right. \\\\\\\left \{ {{\sqrt{6+x} -3\sqrt{3y+4}=-10} \ |\cdot 5 \atop { -5\sqrt{6+x} +4\sqrt{3y+4} =6}} \right. \\\\\\+\left \{ {{5\sqrt{6+x} -15\sqrt{3y+4}=-50} \atop { -5\sqrt{6+x} +4\sqrt{3y+4} =6}} \right. \\----------------\\-11\sqrt{3y+4} =-44\\\\\sqrt{3y+4} =4\\\\(\sqrt{3y+4} )^{2} =4^{2} \\\\3y+4=16\\\\3y=12\\\\\boxed{y=4}\\\\\\\sqrt{6+x} =3\sqrt{3y+4} -10=3\cdot 4-10=12-10=2$

$\displaystyle\bf\\(\sqrt{6+x} )^{2}=2^{2} \\\\6+x=4\\\\\boxed{x=-2}\\\\Otvet \ : \ \Big(-2 \ ; \ 4\Big)$

Answer 2

To solve this system of equations:

1. Let's simplify the given equations: Equation 1: √6 + x - 3√3y + 4 = -10 Equation 2: 4√3y + 4 - 5√6 + x = 6

2. We can rearrange Equation 1 to isolate x and y terms: √6 + x - 3√3y = -14 (subtract 4 from both sides)

3. We can rearrange Equation 2 to isolate x and y terms: x - 5√6 + 4√3y = 2 (subtract 4 from both sides)

4. Now we have two equations: √6 + x - 3√3y = -14 x - 5√6 + 4√3y = 2

5. We can solve this system of equations using the method of substitution or elimination. Let's use substitution:

   From Equation 1, we can solve for x: x = -14 - √6 + 3√3y

   Substitute this expression for x in Equation 2: -14 - √6 + 3√3y - 5√6 + 4√3y = 2

6. Simplify the equation: -14 - √6 - 5√6 + 3√3y + 4√3y = 2

   Combine like terms: -14 - 5√6 - √6 + 7√3y = 2

7. Rearrange the terms: 7√3y - 5√6 - √6 = 2 + 14

   Combine constants: 7√3y - 5√6 - √6 = 16

8. Combine similar radicals: 7√3y - √6(5 + 1) = 16

   Simplify: 7√3y - √6(6) = 16

   7√3y - 6√6 = 16

9. We have the equation: 7√3y - 6√6 = 16

10. Solve this equation for y. Once you have the value of y, substitute it back into either Equation 1 or Equation 2 to solve for x.