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In Mathematics / College | 2025-07-07

Resuelve la siguiente ecuación.
$\frac{3}{4} x+\frac{9}{4}=\frac{1}{7} x-5$

Asked by crisel335555

Answer (1)

Multiply both sides of the equation by 28 to eliminate the fractions: 21 x + 63 = 4 x − 140 .
Subtract 4 x from both sides: 17 x + 63 = − 140 .
Subtract 63 from both sides: 17 x = − 203 .
Divide both sides by 17: x = − 17 203 ​ ​ .

Explanation

Problem Setup We are given the equation 4 3 ​ x + 4 9 ​ = 7 1 ​ x − 5

Eliminating Fractions To eliminate the fractions, we multiply both sides of the equation by the least common multiple of the denominators, which is 28. 28 ( 4 3 ​ x + 4 9 ​ ) = 28 ( 7 1 ​ x − 5 ) Distribute the 28 on both sides: 28 × 4 3 ​ x + 28 × 4 9 ​ = 28 × 7 1 ​ x − 28 × 5 21 x + 63 = 4 x − 140

Isolating x term Subtract 4 x from both sides of the equation: 21 x − 4 x + 63 = 4 x − 4 x − 140 17 x + 63 = − 140

Further isolating x term Subtract 63 from both sides of the equation: 17 x + 63 − 63 = − 140 − 63 17 x = − 203

Solving for x Divide both sides by 17 to solve for x :
x = 17 − 203 ​ x = − 17 203 ​ Since 203 = 17 × 11 + 16 , we can perform long division to see if 203 is divisible by 17. In fact, 203 = 17 × 11 + 16 , so it seems like the fraction cannot be simplified further. However, 17 × 12 = 204 , so 203 is not divisible by 17. Let's try dividing 203 by 17. 203 ÷ 17 = 11.94 , so it's not an integer. Therefore, the fraction is already in its simplest form. We can perform the division to get x = − 11.9411... . However, we can leave the answer as a fraction. x = − 17 203 ​

Final Answer Therefore, the solution to the equation is: x = − 17 203 ​


Examples
When solving problems involving proportions or ratios, you might encounter equations similar to this one. For instance, if you're mixing ingredients for a recipe and need to adjust the quantities to maintain a specific ratio, you might end up with an equation like this. Solving for the unknown variable (x) helps you determine the exact amount of one ingredient needed based on the quantities of the others. This type of algebraic manipulation is fundamental in various fields, including cooking, chemistry, and engineering, where precise measurements are crucial.

Answered by GinnyAnswer | 2025-07-07