How do you solve the following system: #2x-4y=6 , y + 4x = 16 #?

Question

Savannah Anderson · Accepted Answer

See a solution process below:

Step 1) Solve the second equation for #y#: #y + 4x = 16# #y + 4x - color(red)(4x) = 16 - color(red)(4x)# #y + 0 = 16 - 4x# #y = 16 - 4x# Step 2) Substitute #(16 - 4x)# for #y# in the first equation and solve for #x#: #2x - 4y = 6# becomes: #2x - 4(16 - 4x) = 6# #2x - (4 * 16) + (4 * 4x) = 6# #2x - 64 + 16x = 6# #2x + 16x - 64 = 6# #(2 + 16)x - 64 + color(red)(64) = 6 + color(red)(64)# #18x - 0 = 70# #18x = 70# #(18x)/color(red)(18) = 70/color(red)(18)# #x = 35/9# Step 3) Substitute #35/9# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#: #y = 16 - 4x# becomes: #y = 16 - (4 xx 35/9)# #y = 144/9 - 140/9# #y = (144 - 140)/9# #y = 4/9# The Solution Is: #x = 35/9# and #y = 4/9# Or #(35/9, 4/9)#

Hazel Anglin · Answer

undefined To solve the system of equations:

1. Rearrange the equations to isolate one variable in terms of the other:
   
   Equation 1: $2x - 4y = 6$  → $x = 2y + 3$
   
   Equation 2: $y + 4x = 16$ → $x = \frac{{16 - y}}{4}$
   
2. Equate the expressions for $x$ from both equations:
   
   $2y + 3 = \frac{{16 - y}}{4}$
   
3. Solve for $y$:
   
   $8y + 12 = 16 - y$
   
   $9y = 4$
   
   $y = \frac{4}{9}$
   
4. Substitute the value of $y$ into one of the original equations to find $x$:

$x = 2\left(\frac{4}{9}\right) + 3$
   
   $x = \frac{8}{9} + 3$
   
   $x = \frac{35}{9}$

So, the solution to the system is $x = \frac{35}{9}$ and $y = \frac{4}{9}$.