Solve for x: 2(4−3x) + 12 = 7x − 5(2x − 7) - Leaving Cert Mathematics - Question a - 2014

To verify, we substitute $x = -5$ back into the original equation:

Left-hand side:

$2(4 - 3(-5)) + 12 = 2(4 + 15) + 12 = 2(19) + 12 = 38 + 12 = 50$

Right-hand side:

$7(-5) - 5(2(-5) - 7) = -35 - 5(-10 - 7) = -35 + 85 = 50$

Since both sides equal 50, our solution is verified.

Starting with the equations:

1. $x + y = 7$
2. $x^2 + y^2 = 25$

From the first equation, we can express y in terms of x:

$y = 7 - x$

Substituting this expression for y into the second equation:

$(7 - x)^2 + y^2 = 25$

This expands to:

$49 - 14x + x^2 + (7 - x)^2 = 25$

Combining like terms results in:

$49 - 14x + x^2 + 49 - 14x + x^2 = 25$

Simplifying gives:

$2x^2 - 28x + 74 - 25 = 0$
$2x^2 - 28x + 49 = 0$

Using the quadratic formula, we find:

$x = \frac{28 \pm \sqrt{(28)^2 - 4\cdot2\cdot49}}{2\cdot2}$

Calculating the roots: $x = 7 + \sqrt{\frac{25}{2}}$ and $x = 7 - \sqrt{\frac{25}{2}}$ give us the solutions for x and subsequently for y.