Question

X-6=y2 4x+5y 2+3y=26

Answer 1

x-6=y2

x-2y=6

2+3y=26

3y=26-2

3y=24

y=24:3

y=8

Answer 2

It looks like you've provided a system of equations with three equations involving the variables x and y. You seem to have a couple of typos in your equations, so I'll correct them and solve the system for you.

The corrected equations are:

1. $x - 6 = y^2$
2. $4x + 5y^2 + 3y = 26$

Let's solve this system step by step.

Starting with the first equation, $x - 6 = y^2$, we can isolate $x$ in terms of $y$:

$x = y^2 + 6$ (Equation 1)

Now, let's substitute this value of $x$ into the second equation:

$4(y^2 + 6) + 5y^2 + 3y = 26$

Simplify the equation:

$4y^2 + 24 + 5y^2 + 3y = 26$

Combine the like terms:

$9y^2 + 3y + 24 = 26$

Subtract 26 from both sides:

$9y^2 + 3y - 2 = 0$ (Equation 2)

Now, we have a quadratic equation in terms of $y$. To solve for $y$, we can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this equation, $a = 9$, $b = 3$, and $c = -2$. Plugging in these values:

$y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 9 \cdot -2}}{2 \cdot 9}$

Simplifying further:

$y = \frac{-3 \pm \sqrt{9 + 72}}{18}$

$y = \frac{-3 \pm \sqrt{81}}{18}$

$y = \frac{-3 \pm 9}{18}$

This gives us two possible values for $y$:

1. $y = \frac{6}{18} = \frac{1}{3}$
2. $y = \frac{-12}{18} = -\frac{2}{3}$

Now that we have the values of $y$, we can plug them back into Equation 1 to find the corresponding values of $x$:

For $y = \frac{1}{3}$: $x = \left(\frac{1}{3}\right)^2 + 6 = \frac{1}{9} + 6 = \frac{55}{9}$

For $y = -\frac{2}{3}$: $x = \left(-\frac{2}{3}\right)^2 + 6 = \frac{4}{9} + 6 = \frac{58}{9}$

So, the solutions for the system of equations are:

1. $x = \frac{55}{9}$, $y = \frac{1}{3}$
2. $x = \frac{58}{9}$, $y = -\frac{2}{3}$