SOLUTION: I have exams coming up soon but I can't solve these revision questions on simultaneous equations! Can you please explain to me how to solve:
4a+2b=10
9a=27-3b
And
3x-4y=
Algebra ->
Coordinate Systems and Linear Equations
-> Lessons
-> SOLUTION: I have exams coming up soon but I can't solve these revision questions on simultaneous equations! Can you please explain to me how to solve:
4a+2b=10
9a=27-3b
And
3x-4y=
Log On
Question 863301: I have exams coming up soon but I can't solve these revision questions on simultaneous equations! Can you please explain to me how to solve:
4a+2b=10
9a=27-3b
And
3x-4y=10
2x-y=5
Thank you so much if you answer this! I'm trying to be eligible for further maths in my GCSEs. Answer by rothauserc(4718) (Show Source):
You can put this solution on YOUR website! 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
