Calculate the solution to these simultaneous equations using the elimination method: (1)(1)2𝑏−4𝑎2b−4a==−4−4 (2)(2)𝑏+4𝑎b+4a==2222
Question
Calculate the solution to these simultaneous equations using the elimination method: (1)(1)2𝑏−4𝑎2b−4a==−4−4 (2)(2)𝑏+4𝑎b+4a==2222
Solution
Sure, let's solve these equations step by step:
Given equations are: 2b - 4a = -4 ...(1) b + 4a = 22 ...(2)
First, let's simplify equation (1) by dividing it by 2, we get: b - 2a = -2 ...(3)
Now, we have two new equations: b - 2a = -2 ...(3) b + 4a = 22 ...(2)
Now, let's add equation (3) and equation (2), we get: b - 2a + b + 4a = -2 + 22 2b + 2a = 20 2b = 20 - 2a b = 10 - a ...(4)
Now, substitute equation (4) into equation (2), we get: 10 - a + 4a = 22 10 + 3a = 22 3a = 22 - 10 3a = 12 a = 12 / 3 a = 4
Substitute a = 4 into equation (4), we get: b = 10 - 4 b = 6
So, the solution to the system of equations is a = 4 and b = 6.
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