$\frac{3}{4} x+3-2 x=-\frac{1}{4}+\frac{1}{2} x+5$
Answer (1)
Combine like terms on both sides of the equation: − 4 5 x + 3 = 2 1 x + 4 19 .
Multiply both sides by 4 to eliminate fractions: − 5 x + 12 = 2 x + 19 .
Isolate x terms: − 7 = 7 x .
Solve for x: x = − 1 . The solution to the equation is − 1 .
Explanation
Problem Analysis We are given the equation 4 3 x + 3 − 2 x = − 4 1 + 2 1 x + 5 Our goal is to solve for x .
Combining Like Terms First, let's combine the x terms on the left side of the equation: 4 3 x − 2 x = 4 3 x − 4 8 x = − 4 5 x So the left side of the equation becomes: − 4 5 x + 3 Now, let's combine the constant terms on the right side of the equation: − 4 1 + 5 = − 4 1 + 4 20 = 4 19 So the right side of the equation becomes: 2 1 x + 4 19 Now our equation is: − 4 5 x + 3 = 2 1 x + 4 19
Eliminating Fractions To eliminate the fractions, we can multiply both sides of the equation by 4: 4 × ( − 4 5 x + 3 ) = 4 × ( 2 1 x + 4 19 ) Distributing the 4 on both sides, we get: − 5 x + 12 = 2 x + 19
Isolating x Terms Now, let's isolate the x terms. Add 5 x to both sides: 12 = 2 x + 5 x + 19 12 = 7 x + 19 Subtract 19 from both sides: 12 − 19 = 7 x − 7 = 7 x
Solving for x Finally, divide both sides by 7 to solve for x :
7 − 7 = x x = − 1
Final Answer Therefore, the solution to the equation is x = − 1 .
Examples
In electrical engineering, you might use linear equations to analyze circuits. For example, you could determine the current flowing through a resistor in a simple circuit. By setting up an equation that relates voltage, current, and resistance, you can solve for the unknown current. This is a fundamental application of linear equations in circuit analysis.
