Easy Algebra problem

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Easy Algebra problem

If you find my previous problem of "CAR RACING PROBLEM" a little difficult, here is an easy one:

1) B + D = 3C

2) D = 3A + C

3) 3B = C + A

FIND D IN TERMS OF A.

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matherapist	Best Answer Use eqns 1 and 3 to eliminate the B term by multiplying eqn 1 by 3 and subtracting eqn 3 giving 3D=8C-A;(eqn 4) Solve for C in eqn 2 [C=D-3A] and substitute This value of C in eqn 4> 3D=8(D-3A)-A This gives an eqn in D and A so solve for D and get D=5A. Using this info , you can solve for B and C in terms of A and get C=2A and B=A. Posted 144 day ago ( 1 ) ( 0 ) Report Abuse

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Answers (3)

math_fanatic

This is just a series of substitutions:

Equation 2 is almost D in terms of A so the only thing you have to do is find C in terms of A and plug it into equation 2.

I started with the first equation:
B + D = 3C => B + 3A + C = 3C (Substitute second equation in for value of D)

In the third equation, I solved for B in terms of A and C:
B = (A + C)/3

Then I plugged that into the equation from the previous step:
B + 3A + C = 3C => (A + C)/3 + 3A + C = 3C => C + A + 9A + 3C = 9C (Multiplied both sides by 3)

Simplifying further you end up with:
10A = 5C => C = 2A

Now you have C in terms of A so you can plug your new C value into equation 2:
D = 3A + C => D = 3A + 2A => D = 5A

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matherapist

Use eqns 1 and 3 to eliminate the B term by multiplying eqn 1 by 3 and subtracting eqn 3 giving
3D=8C-A;(eqn 4)

Solve for C in eqn 2 [C=D-3A] and substitute This value of C in eqn 4> 3D=8(D-3A)-A

This gives an eqn in D and A so solve for D and get D=5A.

Using this info , you can solve for B and C in terms of A and get C=2A and B=A.

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atta

The easiest way to approach this is by starting with the more simpler equation. However, any equation would work fine. This is simply an instaneous representation of system of equations, a topic covered in elementary algebra. The solution to the above system of equations is as follows:

B + D = 3C => D = 3C - B. Substitution into the 2nd equation yields a more simpler means of deriving one of the variable terms:

D = 3A + C=> (3C - B) = 3A + C => -B = 3A - 2C => B = 2C - 3A. After substitution, this simplies the derivation of either one of the variable terms as shown in the last equation:

3B = C + A => 3(2C - 3A) = C + A => 6C - 9A = C + A. By using algebraic methods, this allows the equating of like terms to derive one of the variables:

5C = 10A => C = 2A. => 3)3B = C + A => 3B = A + (2A) => 3B = 2A => B = 2A/3.

Given C = 2A && B = 2A/3, substituting into the first equation yields the following for D:

B + D = 3C => (2A/3) + D = 3(2A) => 2A/3 + D = 6A => D = 6A - 2A/3 => D = 16A/3.

Given B, C, && D, A can be derived in various ways that supposedly yields no solution.

J.C

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