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Solve the equation.

[tex]\[ \frac{3}{4}x + 3 - 2x = -\frac{1}{4} + \frac{1}{2}x + 5 \][/tex]

If necessary:

- Combine like terms
- Apply properties:
- Add
- Subtract
- Multiply
- Divide

To start over:

- Reset


Sagot :

Sure, let's solve the given equation step-by-step:

The given equation is:
[tex]\[ \frac{3}{4} x + 3 - 2 x = -\frac{1}{4} + \frac{1}{2} x + 5 \][/tex]

### Step 1: Combine like terms on both sides of the equation.

First, let's combine the terms involving [tex]\( x \)[/tex] on the left side:

[tex]\[ \frac{3}{4} x - 2 x + 3 \][/tex]

This can be rewritten as:

[tex]\[ \left( \frac{3}{4} - 2 \right)x + 3 \][/tex]

Next, combine the terms involving [tex]\( x \)[/tex] on the right side:

[tex]\[ -\frac{1}{4} + \frac{1}{2} x + 5 \][/tex]

### Step 2: Simplify the coefficients of [tex]\( x \)[/tex].

[tex]\[ \frac{3}{4} - 2 = \frac{3}{4} - \frac{8}{4} = \frac{3 - 8}{4} = -\frac{5}{4} \][/tex]

So, the equation now looks like this:

[tex]\[ -\frac{5}{4} x + 3 = -\frac{1}{4} + \frac{1}{2} x + 5 \][/tex]

### Step 3: Move all terms involving [tex]\( x \)[/tex] to one side and the constant terms to the other side.

Subtract [tex]\(\frac{1}{2} x\)[/tex] from both sides:

[tex]\[ -\frac{5}{4} x - \frac{1}{2} x + 3 = -\frac{1}{4} + 5 \][/tex]

Simplify the left side:

[tex]\[ -\frac{5}{4} x - \frac{2}{4} x = -\frac{7}{4} x \][/tex]

And combine the constants on the right side:

[tex]\[ -\frac{1}{4} + 5 = 4.75 \][/tex]

So, the equation now is:

[tex]\[ -\frac{7}{4} x + 3 = 4.75 \][/tex]

### Step 4: Simplify further by subtracting 3 from both sides.

[tex]\[ -\frac{7}{4} x = 4.75 - 3 \][/tex]

[tex]\[ -\frac{7}{4} x = 1.75 \][/tex]

### Step 5: Solve for [tex]\( x \)[/tex] by isolating [tex]\( x \)[/tex].

Multiply both sides by [tex]\(-4/7\)[/tex]:

[tex]\[ x = 1.75 \times \left(\frac{-4}{7}\right) \][/tex]

### Final result:

[tex]\[ x = -1.4 \][/tex]

So the solution to the equation [tex]\(\frac{3}{4} x + 3 - 2 x = -\frac{1}{4} + \frac{1}{2} x + 5\)[/tex] is:

[tex]\[ x = -1.4 \][/tex]

The process ensures we carefully combined like terms, applied arithmetic properties, and isolated the variable to find the solution.