Question 863301
1)4a+2b=10
9a=27-3b 
solve second equation for a
a = 3 - b/3
substitute for a in first equation
4(3 - b/3) +2b = 10
12 - 4b/3 + 6b/3 = 10
2b/3 = -2
2b = -6
b = -3
substitute for b in first equation
4a-6 = 10
4a = 16
a = 4
solution a = 4, b = -3
2)3x-4y=10
2x-y=5 
solve second equation for y
-y = 5 - 2x
y = 2x - 5 
now substitute for y in first equation
3x - 4(2x - 5) = 10
3x -8x + 20 = 10
-5x = -10
x = 2
substitute for x in first equation
6 - 4y = 10
-4y = 4
y = -1
solution x = 2, y = -1