3a+9b=8b-a 5a-10b=4a-9b+5

What does a and b equal?

3 a + 9 b = 8 b - a

and
5 a - 10 b = 4 a - 9 b + 5 ?????
-----------------------------
simplify row 1
4 a + 1 b = 0
simplify row two
1 a - 1 b = 5
----------------- add the two rows to get rid of b
5 a = 5
so
a = 1
go back and get b
b = -4

--------------------------------
check

3a+9b=8b-a

5a-10b=4a-9b+5
get them in standard order first///

Na+Mb=k
see if that helps

Are you to solve these equations simultaneously? Does the first equation have a constant or is 3a+9b = 8b-a it?

First combine the terms; for example, the first equation would b ecome
3a + 9b = 8b - a
3a + a + 9b -8b
4a+b = 0
Treat equation 2 the same, the solve by substitution or elimination.

To find the values of a and b, we can solve the given system of equations using the method of elimination or substitution. Let's solve it using the method of elimination.

Given equations:
1) 3a + 9b = 8b - a
2) 5a - 10b = 4a - 9b + 5

To eliminate one of the variables, let's multiply equation (1) by -1 and equation (2) by 3 to make the coefficients of 'a' the same in both equations.

Multiply equation (1) by -1:
-1(3a + 9b) = -1(8b - a)
-3a - 9b = -8b + a

Multiply equation (2) by 3:
3(5a - 10b) = 3(4a - 9b + 5)
15a - 30b = 12a - 27b + 15

Now, let's simplify the equations:
-3a - 9b = -8b + a -> (-3a - a) - 9b + 8b = 0
-4a - b = 0 (equation 3)

15a - 30b = 12a - 27b + 15 -> 15a - 12a - 30b + 27b = 15
3a - 3b = 15 -> a - b = 5 (equation 4)

Now we have a system of equations:
3a - b = 0 (equation 3)
a - b = 5 (equation 4)

To solve this system, we can subtract equation (4) from equation (3) to eliminate 'b'.

3a - b - (a - b) = 0 - 5
3a - b - a + b = -5
2a = -5

Now, solve for 'a':
2a = -5
a = -5/2
a = -2.5

Substitute the value of 'a' (-2.5) into equation (4) to find 'b':
-2.5 - b = 5
b = -2.5 - 5
b = -7.5

Therefore, the solution to the system of equations is a = -2.5 and b = -7.5.