Need Help solving Equations for X

PSP Student 76

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Dec 12, 2012
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Hello, I have to solve equations for X and can't figure it out?

1+3y = 5x+7

2y=2x

and

1/4 + 3/2y= X/4 + 1/2

HELP!!
 
trying

Hello, I have to solve equations for X and can't figure it out?

1+3y = 5x+7

2y=2x

and

1/4 + 3/2y= X/4 + 1/2

HELP!!


Taking a stab:

2y=2x

2/2y=2/2x the "2" cxl

y=x?


1/4 + 3/2y=x/4 + 1/2

1/4 + 3/2y = 2x +4 -4
-4/1 + 1/4 + 3/2y=2x
stuck not really understanding how to solve for x with a y variable.
 
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Let's look at a general equation which all 3 of your problems may be written as:

\(\displaystyle a+by=cx+d\) (Since we are solving for x we require \(\displaystyle c\ne0\).

Now, if we are to solve for x, then we want to isolate x on one side of the equation, with everything else on the other. The first thing we need to do is subtract d from both sides (we are "undoing" the addition by its inverse, which is subtraction):

\(\displaystyle a+by-d=cx+d-d\)

Collect like terms, and I want to use the commutative property of addition (\(\displaystyle a+b=b+a\)) to rearrange a bit on the left.

\(\displaystyle by+a-d=cx\)

Okay, now we have c being multiplied by x on the right, so we "undo" this with division. So, divide both sides by c:

\(\displaystyle \dfrac{by+a-d}{c}=\dfrac{cx}{c}\)

Now, notice the c's will divide out or cancel on the right, and let's rearrange the equation with x on the left side:

\(\displaystyle x=\dfrac{by+a-d}{c}\)

And now we have solved for x. Can you apply this method to your problems now?
 
Hello, I have to solve equations for X and can't figure it out?

1+3y = 5x+7

2y=2x

and

1/4 + 3/2y= X/4 + 1/2 You probably mean 1/4 + (3/2)y = x/4 + 1/2

HELP!!
With respect to the third equation, learn what used to be denis's rule, but is now Jeff's rule.

Get rid of fractions first by multiplying both sides of the equation by the product of the numerators.

\(\displaystyle \dfrac{1}{4} + \dfrac{3}{2} * y = \dfrac{x}{4} + \dfrac{1}{2} \implies (2 * 4) * \left(\dfrac{1}{4} + \dfrac{3}{2} * y\right) = (2 * 4) * \left(\dfrac{x}{4} + \dfrac{1}{2}\right) \implies 2 + 12y = 2x + 4 \implies 1 + 6y = x + 2.\)

Otherwise, the only thing I have to add is that you may have thought that you were to find a numerical solution to these equations. There is no unique numerical solution for problems of this type. When it says solve for x, it means to turn an equation in to the form

x = something. The something may be a number or an expression. For these problems, it means an expression.
 
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