Answer:
Let's solve your system by elimination.
2a + 3c =66 ; 3a + 5c = 105
Multiply the first equation by 5,and multiply the second equation by -3.
5 (2a + 3c = 66)
−3 (3a + 5c = 105)
Becomes:
10a + 15c = 330
−9a − 15c = −315
Add these equations to eliminate c:
a = 15
Now that we've found a let's plug it back in to solve for c.
Write down an original equation:
2a + 3c = 66
Substitute 15 for a in 2a + 3c = 66:
(2)(15) + 3c = 66
3c + 30 = 66 (Simplify both sides of the equation)
3c + 30 −30 = 66 −30 (Add -30 to both sides)
3c = 36
3c3= 363 (Divide both sides by 3)
c = 12
Answer:
a = 15 and c = 12