SOLUTION: My daughter and I(parent) are having problems with her algebra. Solving Equations with the Variable on each side. Would please help with a few problems so that we may be able to un

Question 19353: My daughter and I(parent) are having problems with her algebra. Solving Equations with the Variable on each side. Would please help with a few problems so that we may be able to understand a little better. She has to justify her answer.
1. 3/8 - 1/4t = 1/2t - 3/4
2. 20c + 5 = 5c + 65
3. 7 - 3r = r - 4(2+r)
Thank you for your help.
Found 2 solutions by venugopalramana, Earlsdon:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
in these problems ,transfer all known to one side of the = sign ,usually rhs and all unknowns to the other side of = sgn ,usually lhs of = sign...while transferring from one side of = to the other + becomes - and - becomes +...now simplify both sides...now you will get finally (some number)*(unknown)=known value..now the unknown =known value/ (some number)..which was multiplying the unknown earlier on the lhs of = sign..SEE THE EXAMPLES BELOW..
1. 3/8 - 1/4t = 1/2t - 3/4
3/8+3/4=1/2t+1/4t...(consider t as some apple ..this means add 1/2 apple to 1/4 apple..we keep aside the apple and add the numbers )=t(1/2+1/4)=t((2+1)/4)
=t(3/4)...so we got
t(3/4)=3/8+3/4=((3+2*3)/8=(3+6)/8=9/8
t(3/4)=9/8
t=(9/8)/(3/4)=(9*4)/(3*8)=3/2...
so t=3/2
you can verify the answer by substituting for t the value of 3/2 and check whether the equation is satisfied..we get ...
3/8-(1/4)*(3/2)=3/8-3/8=0 is the lhs of =
(1/2)*(3/2)-3/4=3/4-3/4=0 is the rhs of =..
so lhs = rhs ..so our answer is correct.
2. 20c + 5 = 5c + 65
20c-5c=65-5=60
15c=60
c=60/15=4
c=4...check the answer as before
3. 7 - 3r = r - 4(2+r)
7-3r=r-4*2-4*r
7-3r=r-8-4r
7+8=15=r-4r+3r=r(1-4+3)=r*0=0...THE PROBLEM DOES NOT HAVE A PROPER SOLUTION
SINCE LHS=15 and RHS=0..THEY CANNOT BE EQUAL
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
The main idea here, of course, is to get the variable onto one side of the equation and the numbers onto the other side. The only restriction is, you must use only the valid rules of algebra.
1) Add to both sides of the equation. Justification...whatever you do to one side, you must do exactly the same to the other side.
Now simplify this.
Add to both sides. Same justification as above.
Simplify.
Finally, multiply both sides by the multiplicative inverse of 3/4...that's 4/3. Justification...a number multiplied by its multiplicative inverse equals 1 and this leaves you with 1 X t or just t on the right side.
Simplify.
or
2)
Subtract 5c from both sides. To get the variables together on one side of the equation.
Subtract 5 from both sides. To get the numbers onto one side of the equation.
Finally, divide both sides by 15. To get c by itself.

3)
Use the distributive property to expand the right side.
Collect like-terms on the right side.
Add 3r to both sides. To get the variable onto one side of the equation.
This can't be true. So the initial equation was inconsistent, i.e., not really an equation.

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