can someone help me with solving this?

5/2x+1/4x=9/4+x

Solve it like any other equation - just add and subtract fractions.

(5/2)x + (1/4)x = (9/4) + x
(11/4)x = (9/4) + x
(7/4)x = (9/4)

I was collecting the like terms first and I ended up with 11/4x=9/4+x then I was stuck.

I ended up with this as the answer
9/7 though

Remember that x = (1/1)x = (4/4)x.

x = 9/7 is correct.

Of course! I'd be happy to help you solve the equation. To begin, let's simplify the equation step by step.

The given equation is: 5/(2x) + 1/(4x) = 9/4 + x

To solve this equation, we need to get rid of the fractions. To eliminate the fractions, we can find a common denominator for the fractions on the left side of the equation. The least common denominator (LCD) for 2x and 4x is 4x. Multiply each term on the left side by 2 to eliminate the denominators:

(2)(5)/(2x) + (2)(1)/(4x) = (2)(9)/4 + (2)(x)

Simplifying this expression gives us:

10/x + 2/x = 18/4 + 2x

Now, we can combine the fractions on the left side:

(10 + 2)/x = 18/4 + 2x

Simplifying the right side yields:

12/x = 18/4 + 2x

Next, let's simplify the right side by converting the fraction to a decimal:

18/4 = 4.5

Our equation becomes:

12/x = 4.5 + 2x

To remove the denominator, multiply both sides of the equation by x:

12 = 4.5x + 2x^2

Now, we have a quadratic equation. Let's rearrange it in standard form:

2x^2 + 4.5x - 12 = 0

To solve this quadratic equation, we can factor, complete the square, or use the quadratic formula. I'll use the quadratic formula in this case:

x = (-b ± √(b^2 - 4ac)) / (2a)

In our equation, a = 2, b = 4.5, and c = -12. Substituting these values into the formula, we get:

x = (-(4.5) ± √((4.5)^2 - 4(2)(-12))) / (2(2))

Simplifying further:

x = (-4.5 ± √(20.25 + 96)) / 4

x = (-4.5 ± √116.25) / 4

The square root of 116.25 is 10.77 (rounded to two decimal places). Now we can find the two possible solutions for x:

x1 = (-4.5 + 10.77) / 4

x1 = 6.27 / 4

x1 ≈ 1.57

x2 = (-4.5 - 10.77) / 4

x2 = -15.27 / 4

x2 ≈ -3.82

Therefore, the solutions for the given equation are approximately x ≈ 1.57 and x ≈ -3.82.