Solve the simultaneous equations - Leaving Cert Mathematics - Question 1 - 2018

To solve the system of equations, we can use substitution or elimination.

1. From the first equation, isolate z: $z = 2x + 3y + 4$

2. Substitute this expression for z into the other two equations:

   • For the second equation: $3x + 2y + 2(2x + 3y + 4) = 14$ Simplifying this gives:
   7x + 8y + 8 = 14\ 7x + 8y = 6 \\$$ - For the third equation: $$x - 3(2x + 3y + 4) = -13\ x - 6x - 9y - 12 = -13\ -5x - 9y = -1 \\$$ Therefore, we have two new equations: 1. $7x + 8y = 6$ 2. $-5x - 9y = -1$

3. Solving these two equations simultaneously will yield the values for x and y:

   • Multiply the first equation by 5:
   -5x - 9y = -1$$ Now we can add these equations to eliminate x: $$35x + 40y - 5x - 9y = 30 - 1\ 30x + 31y = 29\ y = \frac{29 - 30x}{31}$$ Substitute y back to one of the original equations to find x.

4. Calculating this leads to x = 2 and when substituted back, y = -1 and then find z using initial substitution.