Find three consecutive odd integers such that four times the middle integer is two more than the sum of the first and third.
I think I'm doing it wrong.
4((x)+(x+2)=x+(x+4)+2
4(2x+4)=2+3x+6
8x+16=3x+8
5x=8 ?
Please help
A real estate agent sold two homes and received commissions totaling $6000. The agent's commission on one home was one and one-half times the commission on the second home. If p represents the smaller commission, write an equation to determine the agent's commission on each home.
P= smaller number
6000-P= Larger number
I don't get how to write it as an equation
To solve this problem correctly, let's go through the steps again:
1. Let's start by representing the first odd integer as "x". Since we are looking for consecutive odd integers, the next two odd integers will be "x + 2" and "x + 4".
2. According to the information in the problem, four times the middle integer (4 * (x + 2)) is two more than the sum of the first and third integers (x + (x + 4) + 2).
So the equation becomes: 4 * (x + 2) = (x + x + 4) + 2.
3. Simplify the equation:
4x + 8 = 2x + 6 + 2.
4x + 8 = 2x + 8.
4. Subtract 2x from both sides to isolate the variable:
4x - 2x + 8 = 8.
5. Combine like terms:
2x + 8 = 8.
6. Subtract 8 from both sides:
2x = 0.
7. Divide both sides by 2:
x = 0 / 2.
x = 0.
So, the first odd integer is 0.
To find the other consecutive odd integers, substitute the value of x back into the expression:
0 + 2 = 2, and
0 + 4 = 4.
Therefore, the three consecutive odd integers are 0, 2, and 4.