How do you solve #\frac { n } { 2} + \frac { n - 1} { 2} = 1#?

Question

Avery Alexander · Accepted Answer

See a solution process below:

First, add the fractions on the left side of the equation by adding the numerators over the common denominator: #(n + (n - 1))/2 = 1# #(1n + 1n - 1)/2 = 1# #(2n - 1)/2 = 1# Next, multiply each side of the equation by #color(red)(2)# to eliminate the fraction while keeping the equation balanced: #color(red)(2) xx (2n - 1)/2 = color(red)(2) xx 1# #cancel(color(red)(2)) xx (2n - 1)/color(red)(cancel(color(black)(2))) = 2# #2n - 1 = 2# Then, add #color(red)(1)# to each side of the equation to isolate the #n# term while keeping the equation balanced: #2n - 1 + color(red)(1) = 2 + color(red)(1)# #2n - 0 = 3# #2n = 3# Now, divide each side of the equation by #color(red)(2)# to solve the equation for #n# while keeping the equation balanced: #(2n)/color(red)(2) = 3/color(red)(2)# #(color(red)(cancel(color(black)(2)))n)/cancel(color(red)(2)) = 3/2# #n = 3/2#

Eva Adamson · Answer

undefined To solve the equation $ \frac{n}{2} + \frac{n - 1}{2} = 1 $, follow these steps:

1. Combine the fractions on the left side.
2. Solve for $ n $.

$$ \frac{n}{2} + \frac{n - 1}{2} = 1 $$

Combine fractions:
$$ \frac{n + (n - 1)}{2} = 1 $$

$$ \frac{2n - 1}{2} = 1 $$

Multiply both sides by 2 to clear the fraction:
$$ 2n - 1 = 2 $$

Add 1 to both sides:
$$ 2n = 3 $$

Divide both sides by 2:
$$ n = \frac{3}{2} $$

So, the solution to the equation is $ n = \frac{3}{2} $.