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Solve for 
x.

(4x+2)/(3)=(x+3)/(2)
Answer:

Solve for x \mathrm{x} .\newline4x+23=x+32 \frac{4 x+2}{3}=\frac{x+3}{2} \newlineAnswer:

Full solution

Q. Solve for x \mathrm{x} .\newline4x+23=x+32 \frac{4 x+2}{3}=\frac{x+3}{2} \newlineAnswer:
  1. Identify Equation: Identify the equation to solve for xx. We have the equation 4x+23=x+32\frac{4x+2}{3} = \frac{x+3}{2}. We need to find the value of xx that makes this equation true.
  2. Cross-Multiply: Cross-multiply to eliminate the fractions.\newlineTo get rid of the fractions, we can cross-multiply. This means we multiply both sides of the equation by the denominators of the other side.\newline(4x+2)×2=(x+3)×3(4x + 2) \times 2 = (x + 3) \times 3
  3. Distribute Multiplication: Distribute the multiplication over addition on both sides.\newlineNow we distribute the multiplication over addition on both sides of the equation.\newline(4x×2)+(2×2)=(x×3)+(3×3)(4x \times 2) + (2 \times 2) = (x \times 3) + (3 \times 3)\newline8x+4=3x+98x + 4 = 3x + 9
  4. Move Terms: Move all xx terms to one side and constant terms to the other side.\newlineWe want to isolate xx on one side of the equation. To do this, we subtract 3x3x from both sides and subtract 44 from both sides.\newline8x3x+44=3x3x+948x - 3x + 4 - 4 = 3x - 3x + 9 - 4\newline5x=55x = 5
  5. Divide Coefficient: Divide both sides by the coefficient of xx to solve for xx. To isolate xx, we divide both sides of the equation by 55. 5x5=55\frac{5x}{5} = \frac{5}{5} x=1x = 1

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