Solve for $x$: $$3(x - 7) + 5(x - 4) = 15$$, where $x \in \mathbb{R}$ - Leaving Cert Mathematics - Question 3 - 2016

We have two equations:

1. $4a + 3b = -3$
2. $5a = 25 + 2b$

For the second equation, rearrange it to isolate $b$:

$5a - 25 = 2b$ $b = \frac{5a - 25}{2}$

Now, substitute this expression for $b$ into the first equation:

$4a + 3\left(\frac{5a - 25}{2}\right) = -3$

Multiplying through by 2 to clear the fraction gives:

$8a + 3(5a - 25) = -6$

Expanding results in:

$8a + 15a - 75 = -6$ $23a - 75 = -6$

Adding 75 to both sides results in:

$23a = 69$

Now, divide by 23:

$a = 3$

Now, substitute $a = 3$ back into the equation for $b$:

$b = \frac{5(3) - 25}{2}$ $b = \frac{15 - 25}{2} = \frac{-10}{2} = -5$

Thus, the values are $a = 3$ and $b = -5$.

First, expand the inequality:

$2(2x - 3) + 6x < 25$

This gives:

$4x - 6 + 6x < 25$

Combining like terms results in:

$10x - 6 < 25$

Adding 6 to both sides yields:

$10x < 31$

Dividing by 10 results in:

$x < 3.1$

Since $x$ must be a natural number ($x \in \mathbb{N}$), the possible values of $x$ are:

$x = 1, 2, 3$