Solving the Equation: (3x-2)(x+5) = (x+2)(x+1)
This article will guide you through the steps to solve the equation (3x-2)(x+5) = (x+2)(x+1).
Expanding the Equation
Firstly, we need to expand both sides of the equation by using the distributive property (also known as FOIL):
- Left side: (3x-2)(x+5) = 3x(x+5) - 2(x+5) = 3x² + 15x - 2x - 10 = 3x² + 13x - 10
- Right side: (x+2)(x+1) = x(x+1) + 2(x+1) = x² + x + 2x + 2 = x² + 3x + 2
Now our equation looks like this: 3x² + 13x - 10 = x² + 3x + 2
Simplifying the Equation
Next, let's move all the terms to one side to set the equation equal to zero:
- Subtract x² from both sides: 2x² + 13x - 10 = 3x + 2
- Subtract 3x from both sides: 2x² + 10x - 10 = 2
- Subtract 2 from both sides: 2x² + 10x - 12 = 0
Solving the Quadratic Equation
We now have a quadratic equation in the form ax² + bx + c = 0. We can solve this using the quadratic formula:
x = [-b ± √(b² - 4ac)] / 2a
In our equation:
- a = 2
- b = 10
- c = -12
Substitute these values into the quadratic formula:
x = [-10 ± √(10² - 4 * 2 * -12)] / (2 * 2) x = [-10 ± √(100 + 96)] / 4 x = [-10 ± √(196)] / 4 x = [-10 ± 14] / 4
This gives us two possible solutions:
- x1 = (-10 + 14) / 4 = 1
- x2 = (-10 - 14) / 4 = -6
Conclusion
Therefore, the solutions to the equation (3x-2)(x+5) = (x+2)(x+1) are x = 1 and x = -6.