Question 14 (a) (i) Factorise $x^2 + 6x - 7$ - Junior Cycle Mathematics - Question 14 - 2016

Now, using the factors previously found, we can set each factor to zero:

1. From $(x + 7) = 0$: $x = -7$

2. From $(x - 1) = 0$: $x = 1$

Therefore, the solutions to the equation $x^2 + 6x - 7 = 0$ are: $x = -7\ ext{ and } x = 1$

To solve the simultaneous equations, we can start with the second equation:

$x + 2y = 25 \implies x = 25 - 2y$

Next, we can substitute this expression for $x$ into the first equation:

$3(25 - 2y) + 2y = 39$

Expanding this gives: $75 - 6y + 2y = 39$

Combining like terms results in: $75 - 4y = 39$

Now, isolating $y$: $-4y = 39 - 75$ $-4y = -36$ $y = 9$

Now substituting $y = 9$ back into $x = 25 - 2y$: $x = 25 - 2(9) = 25 - 18 = 7$

Thus, the solution to the simultaneous equations is: $x = 7\ ext{ and } y = 9$