Question

3y + 2 = 4x , x+ 5y = 3 what is the solution to the system of equations shown above? (in fraction form)

Answer 1

Answer:

19/23, 50/23

Step-by-step explanation:

We can take the first equation and isolate the x:

x + 5y = 3 ⇒ -5y ⇒

x = 3 - 5y.

Then we can substitute the x in the first equation by this:

3y + 2 = 4(3-5y) ⇒

3y + 2 = 12 - 20y ⇒ +20y

23y + 2 = 12 ⇒ -2

23y = 10 ⇒ ÷23

y = $\frac{10}{23}$

Then we can add the value of y into the equation equaling x we made:

x = 3 - 5($\frac{10}{23}$) ⇒

x = 3 - $\frac{50}{23}$ ⇒

x = $\frac{19}{23}$

Hope this helps! :)

Answer 2

We have x+5y=3 => x= 3-5y
So 3y+2=4x
<=>3y+2=4(3-5y)
<=>3y+2=12-20y
<=>3y+20y=12-2
<=>23y=10
<=>y=10/23 => x=3-5(10/23)=19/23

Ok done. Thank to me :>