How do you solve the system of linear equations #x + 3y = 5# and #2x - y = 5#?

Question

Avery Alexander · Accepted Answer

#20/7 = x#, #5/7 = y# #x+3y = 5# #2x - y = 5# If both equations equal #5#, we can set them equal to eachother #x + 3y = 2x - y# #4y = x# Now we can substitute #4y# for #x# in one of the equations (let's pick the first one) #4y + 3y = 5# #7y = 5# #y = 5/7# Now, if #4y = x#, then #4 xx 5/7 = x# or #x = 20/7# Now to check our work. Let's substitute #5/7# and #20/7# for #y# and #x# in the second equation. If we have the correct answers, our equation should still equal #5#. #2(20/7) - 5/7# #40/7 - 5/7# #35/7#, which simplifies to #5#! So we were correct.

Abigail Allen · Answer

undefined To solve the system of linear equations $x + 3y = 5$ and $2x - y = 5$:

1. **Method 1: Substitution**
   Solve one equation for one variable and substitute into the other equation:
   From $2x - y = 5$, solve for $y$:
   $$y = 2x - 5$$
   Substitute this into $x + 3y = 5$:
   $$x + 3(2x - 5) = 5$$
   Simplify:
   $$x + 6x - 15 = 5$$
   $$7x - 15 = 5$$
   $$7x = 20$$
   $$x = \frac{20}{7}$$

Substitute $x$ back into $y = 2x - 5$:
   $$y = 2\left(\frac{20}{7}\right) - 5$$
   $$y = \frac{40}{7} - \frac{35}{7}$$
   $$y = \frac{5}{7}$$

So, the solution is $x = \frac{20}{7}$ and $y = \frac{5}{7}$.

2. **Method 2: Elimination**
   Multiply the second equation by 3 to eliminate $y$:
   $$3(2x - y) = 3(5)$$
   $$6x - 3y = 15$$

Now, we have the system:
   $$\begin{cases} x + 3y = 5 \ 6x - 3y = 15 \end{cases}$$

Add the equations together to eliminate $y$:
   $$(x + 3y) + (6x - 3y) = 5 + 15$$
   $$7x = 20$$
   $$x = \frac{20}{7}$$

Substitute $x$ into $x + 3y = 5$:
   $$\frac{20}{7} + 3y = 5$$
   $$3y = 5 - \frac{20}{7}$$
   $$3y = \frac{35}{7} - \frac{20}{7}$$
   $$3y = \frac{15}{7}$$
   $$y = \frac{5}{7}$$

So, the solution is $x = \frac{20}{7}$ and $y = \frac{5}{7}$.