Impossible to Solve? (Easy)

Algebra Level 1

When solving systems of linear equations, we are taught that we need just as many equations as there are variables to solve the system. However, you don't need to have all the equations if you are solving for one variable.. Here's an easy system of two linear equations with three variables, yet you can solve one of the variables. Give it a try!

If 2 a + 3 b + 2 c = 12 2 a b 2a+3b+2c=12-2a-b and c = a + b c=a+b , then c = c=

4 2 3 1

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1 solution

Zyberg Nee
Jan 7, 2016

Personally I prefer to solve this kind of problems without pencils or any drawings as I find it to be a little faster.

Let's simplify first equation a little bit:

2 a + 3 b + 2 c = 12 2 a b 2a+3b+2c=12-2a-b

2 a + 2 a + 3 b + b + 2 c = 12 2a+2a+3b+b+2c=12

4 a + 4 b + 2 c = 12 4a+4b+2c=12 we can divide everything by 2 2 .

2 a + 2 b + c = 6 2a+2b+c=6 we know from the problem that c = a + b c=a+b

2 ( a + b ) + c = 6 2(a+b)+c=6

2 c + c = 6 2c+c=6

3 c = 6 3c=6

c = 2 c=2

I got the answer, but I solved it using systems of equations. It just so happens that when I multiplied 'c=a+b' by 4 , I was able to eliminate the variables 'a' and 'b' at the same time. this left only 'c' to solve. I truly need to force myself to think outside of the 'box'

Paul Alston - 2 months, 1 week ago

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