1 1 7
-y - -= -
2 8 8
please help me!
It isn't clear to me what this expression means. Can you clarify it?
(one-half times "y") minus one-eighths equals seven-eighths
or
(1/2 * y)- 1/8= 7/8
solve the equation, then check your solution.
Ok, this is how I would've written that
(1/2)y - 1/8 = 7/8
Add 1/8 to both sides, then tell us what you would do to have the y by itself on the lefthand side.
what? can you just tell me how to do the problem? i know how to check it.:) thx
The basics of algebra are that if you'e given an equation you can add the same quantity to both sides and the equality is unaffected. You can also multiply both sides by the same quantity and the equality remains the same.
In your equation what you're trying to solve for is y, so you want to work the numbers to the righthand side and leave y by itself on the lefthand side.
If you add 1/8 to both sides that leaves (1/2)y on the lefthand and 1 on the righthand. What should you multiply both sides by to have y by itself on the lefthand side?
If I just tell you how to do the problem without you understanding the steps, then what have you learned?
I apologize for the confusion. Let me break it down for you.
Given the equation: (1/2)y - 1/8 = 7/8
To solve for y, you want to isolate y on one side of the equation. Start by adding 1/8 to both sides of the equation:
(1/2)y - 1/8 + 1/8 = 7/8 + 1/8
Simplifying:
(1/2)y = 8/8
Now, you need to get y by itself on the left side of the equation. Since (1/2) is multiplying y, you can divide both sides by (1/2):
(1/2)y / (1/2) = (8/8) / (1/2)
Simplifying:
y = 8/8 / 1/2
To divide fractions, multiply the first fraction by the reciprocal of the second fraction:
y = (8/8) * (2/1)
Simplifying:
y = 16/8
Reducing the fraction to its simplest form:
y = 2
So, the solution to the equation is y = 2.
Now, you can check your solution by substituting y = 2 back into the original equation:
(1/2)(2) - 1/8 = 7/8
Simplifying:
1 - 1/8 = 7/8
8/8 - 1/8 = 7/8
7/8 = 7/8
The equation holds true, which confirms that the solution y = 2 is correct.