Substitute the value of $x$ back into the equation $\frac{3}{4} x+3 -2 x=-\frac{1}{4}+\frac{1}{2} x+5$ to justify the solution. What value of $x$ is the solution?
$x=\square$
Answer (1)
Combine like terms in the equation 4 3 x + 3 − 2 x = − 4 1 + 2 1 x + 5 .
Isolate x on one side of the equation to get − 4 7 x = 4 7 .
Solve for x , which gives x = − 1 .
Substitute x = − 1 back into the original equation to verify the solution: 4 17 = 4 17 .
The solution is − 1 .
Explanation
Solving for x First, we need to solve the equation for x . The given equation is: 4 3 x + 3 − 2 x = − 4 1 + 2 1 x + 5 We will combine like terms to isolate x .
Isolating x Combine the x terms on the left side: 4 3 x − 2 x = 4 3 x − 4 8 x = − 4 5 x So the equation becomes: − 4 5 x + 3 = − 4 1 + 2 1 x + 5 Now, subtract 3 from both sides: − 4 5 x = − 4 1 + 2 1 x + 2 Convert 2 to a fraction with denominator 4: 2 = 4 8 So the equation becomes: − 4 5 x = − 4 1 + 2 1 x + 4 8 Combine the constants on the right side: − 4 5 x = 4 7 + 2 1 x Subtract 2 1 x from both sides. Note that 2 1 x = 4 2 x :
− 4 5 x − 4 2 x = 4 7 − 4 7 x = 4 7 Multiply both sides by − 7 4 to solve for x :
x = 4 7 × − 7 4 x = − 1
Verifying the Solution Now we need to substitute x = − 1 back into the original equation to verify the solution: 4 3 x + 3 − 2 x = − 4 1 + 2 1 x + 5 4 3 ( − 1 ) + 3 − 2 ( − 1 ) = − 4 1 + 2 1 ( − 1 ) + 5 − 4 3 + 3 + 2 = − 4 1 − 2 1 + 5 − 4 3 + 5 = − 4 1 − 4 2 + 5 − 4 3 + 5 = − 4 3 + 5 4 − 3 + 20 = 4 − 3 + 20 4 17 = 4 17 The left side equals the right side, so the solution x = − 1 is correct.
Final Answer Therefore, the solution to the equation is x = − 1 .
Examples
In electrical engineering, you might use a similar equation to calculate the current in a circuit. The variable x could represent the current, and the fractions and constants could represent resistances and voltage sources. Solving the equation would tell you the current flowing through the circuit. Verifying the solution by substituting it back into the original equation ensures that your calculations are correct and that the circuit will behave as expected. This is crucial for designing safe and efficient electrical systems.
