3a + 5b = 70
6a + 9b = 135
Multiply both sides of the 1st Eq by -6.
Multiply both sides of the 2nd Eq by 3.
Then add the Eqs.
-18a - 30b = -420
+18a + 27b = 405
-3b = -15
b = 5.
in the 1st Eq, replace b by 5:
3a + 5*5 = 70
3a = 70-25 = 45
a = 15
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To solve this system of equations for the variables 'a' and 'b', we can use the method of substitution or elimination. In this case, I will use elimination to solve the system.
First, we need to make the coefficients of either 'a' or 'b' the same in both equations. Let's eliminate the 'a' term by multiplying the first equation by 2 and the second equation by 1:
(2)(3a + 5b) = (2)(70) -> 6a + 10b = 140
(1)(6a + 9b) = (1)(135) -> 6a + 9b = 135
Now, we can subtract the second equation from the first equation:
(6a + 10b) - (6a + 9b) = 140 - 135
6a - 6a + 10b - 9b = 5
b = 5
Now that we have found the value of 'b', we can substitute it back into either equation to solve for 'a'. Let's use the first equation:
3a + 5(5) = 70
3a + 25 = 70
3a = 70 - 25
3a = 45
a = 45 / 3
a = 15
Therefore, the solution to the system of equations is a = 15 and b = 5.