What is the solution to this system of linear equations? 12q + 3r = 15 –4q – 4r = –44 (–18, 29) (–2, 13) (8, –1) (15, –44)

12q +3r =15    
3(4q +r) = 3*5  
Divide both sides by 3
4q +r = 5
Time for the second one:
-4q -4r = -44  
-4(q +r) = (-4)*11  
Divide both sides by -4
 q +r = 11
4q +r = 5
 q +r = 11
Subtract the second from first equation 
3q 0 = -6
3q = -6
q = -6/3 = -2
q = -2
q +r = 11
-2 +r =11
r = 11 +2 
r = 13 
q = -2
r = 13
Your answer is...
(-2,13)

Since we want to solve for the variable x, we want to isolate x

a²x + (a - 1) = (a + 1)x ⇒ Distribute x to (a+1). Also, remove parentheses

a²x + a - 1 = ax + x ⇒ Subtract a from both sides

a²x - 1 = ax + x - a ⇒ Add 1 to both sides

a²x = ax + x - a + 1 ⇒ Subtract (ax + x) from both sides

a²x - (ax + x)= ax + x - a + 1 - (ax+x) ⇒ Simplify. Remember that multiplying positive by negative = negative

a²x - ax - x = ax + x - a + 1 - ax - x ⇒ Simplify

a²x - ax - x = -a + 1 ⇒ Factor out the x from a²x - ax - x

x(a² - a - 1) = -a + 1 ⇒ Divide both sides by (a² - a - 1)

x = (-a + 1) / (a² - a - 1)

However, we need to make sure that the denominator does not equal 0. Therefore, you set the denominator = 0 (just use the quadratic formula for this), and it gives that the denominator =0 when a = (1+√5)/2 AND (1-√5)/2

Therefore, the final answer is

x = (-a + 1) / (a² - a - 1) given that a ≠ (1+√5)/2, a ≠ (1-√5)/2