How do you solve the system #1/4x+y=3/4# and #3/2x-y=-5/2#?

Question

Isaiah Anthony · Accepted Answer

From the first equation we can derive a formula for #y# by subtracting #1/4x# from both sides to get

#y = 3/4 - 1/4x#

Then substitute that for #y# in the second equation:

#-5/2 = 3/2x-y#

#= 3/2x-(3/4-1/4x)#

These fractions are a pain, so let's eliminate them by multiplying through by 4 to get:

#-10=6x-(3-x) = 7x-3#

Now add 3 to both sides to get

#7x = -7#

Divide both sides by #7# to get

#x = -1#

Then #y = 3/4 - 1/4x = 3/4 + 1/4 = 1#

Eli Assouline · Answer

undefined To solve the system $ \frac{1}{4}x + y = \frac{3}{4} $ and $ \frac{3}{2}x - y = -\frac{5}{2} $:

1. Solve one of the equations for one variable.
2. Substitute the expression found in step 1 into the other equation.
3. Solve for the remaining variable.
4. Once you have one variable's value, substitute it back into one of the original equations to find the other variable's value.

Let's solve it:

1. From the first equation:
   $ \frac{1}{4}x + y = \frac{3}{4} $
   $ y = \frac{3}{4} - \frac{1}{4}x $

2. Substitute $ y = \frac{3}{4} - \frac{1}{4}x $ into the second equation:
   $ \frac{3}{2}x - (\frac{3}{4} - \frac{1}{4}x) = -\frac{5}{2} $

3. Solve for $ x $:
   $ \frac{3}{2}x - \frac{3}{4} + \frac{1}{4}x = -\frac{5}{2} $
   $ \frac{7}{4}x - \frac{3}{4} = -\frac{5}{2} $
   $ \frac{7}{4}x = -\frac{5}{2} + \frac{3}{4} $
   $ \frac{7}{4}x = -\frac{10}{4} + \frac{3}{4} $
   $ \frac{7}{4}x = -\frac{7}{4} $
   $ x = -1 $

4. Substitute $ x = -1 $ into the first equation to find $ y $:
   $ \frac{1}{4}(-1) + y = \frac{3}{4} $
   $ -\frac{1}{4} + y = \frac{3}{4} $
   $ y = \frac{3}{4} + \frac{1}{4} $
   $ y = 1 $

Therefore, the solution to the system is $ x = -1 $ and $ y = 1 $.