Help with these two math questions, please?

Question

I need help with these two math problems..I thought I knew how to do the first one but I was wrong

Please solve for the following variable for these two questions:

1) 4 ( 1/2x - 1) = 2 ( x - 2 )

2) { x+y = 5

{3x-y=3

Answer 1

1)  4 ( 1/2x - 1) = 2 ( x - 2 )

2x - 4 = 2x - 4

therefore there are infinite solutions. thats the answer to #1

2)  x+y = 5            

    3x-y=3

the "y's" cancel out because of the fact

that 1's positive and the other is negative so 4x=8. the end solution would be    x=2

Answer 2

4 ( 1/2x - 1 ) = 2 ( x - 2 )

2x - 4 = 2x - 4

1 = 1

x can be anything, its called an identity.

x + y = 5

3x - y = 3

solve for x in first equation

x = 5 - y

put it into the 2nd equation

3 ( 5 - y ) - y = 3

15 - 3y - y = 3

-4y = -12

y = 3

put that back into the first equation

x + 3 = 5

x = 2

Answer 3

Question 1:

4(1/2x - 1) = 2 (x - 2)

2x - 4 = 2x - 4

2x = 2x - 4 + 4

2x = 2x

x = 1

????? ok i got myself confused with this question too...what's the answer by the way?

Question 2:

x + y = 5...equation 1

3x - y = 3...equation 2

take eq.1 (x + y = 5);

y = 5 - x...equation 3

replace eq. 3 in eq. 2;

3x - (5 - x) = 3

3x - 5 + x = 3

4x = 3 + 5

4x = 8

x = 2

take x and replace it in equation 1;

2 + y = 5

y = 5 - 2

y = 3

therefore; x = 2,y = 3

Answer 4

Green Eyes here it is:

1) 4 ( 1/2x - 1) = 2 ( x - 2 )

4/2(1/2x - 1) = (x - 2)

(x - 4) = (x - 4)

No solution

2) { x+y = 5

{3x-y=3

x + y = 5 ← eq 1

3x - y = 3 ← eq 2

add equation 1 & 2

4x = 8

x = 2

Substitute x=2 to eq 1

2 + y = 5

y = 5 - 2

y = 3

=================================

Ans:: x = 2 & y = 3

=================================

hope this helps

Answer 5

1.)  4(1/2x - 1)= 2(x-2)

2x - 4 = 2x - 4

0 = 0

There is no explicit solution for x. It is part of the Real Number set:

x ε (R)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

2.)    x + y = 5

       3x -y = 3

3x + 3y = 15  Multiply both sides of first equation by 3

-3x +  y  = 3   Subtract second equation from first

¯¯¯¯¯¯¯¯¯¯¯¯

     4y = 12

      y = 3

     ¯¯¯¯¯

    

Answer 6

2x - 4 = 2x - 4

0 = 0

x = any real or imaginary number

y = 5 - x

3x - 5 + x = 3

4x = 8

x = 2

Answer 7

1. 4(1/2x - 1) = 2(x - 2)

First distibute on both sides

2x - 4 = 2x - 4

Since the left and right are ALWAYS equal no matter what x you plug in...

Answer:  All real numbers

2.  Here's a good explanation...

http://www.mathclues.com/MSubstitutionMe...

Hope this helps you!  Please email anytime for further questions.

Answer 8

1)  2x - 4 = 2x - 4

    0=0

    therefore no answer/ ALL read numbers!

2) is it like: x=5-y and 3x-y=3 then u sub in the first modified equation in for x: 3(5-y)-y=3 which becomes 15-3y-y=3 then -4y=-12 then y=3.

Hope u get them ~ email me at [e-mail] if you have more questions

Answer 9

Here's what I recommend for finding your mistake.  As you manipulate the equations, copy them into QuickMath's equation solver and have it solve the equation you have created.  When it gives a different answer than the one it gives for the original equation, then you'll know where your mistake is.

Here's how I'd do 1) by hand:

4 ( 1/2x - 1) = 2 ( x - 2 )

(2x - 4) = 2x - 4

Hmm... x can be any value at all.  This is an identity.

QuickMath can do simultaneous equations.  Use the Advanced mode under Equations: Solve.

2)  QuickMath says x=2, y=3