How do you solve the following system: #x-2y=3, 2x+3y =4
#?

Question

How do you solve the following system: #x-2y=3, 2x+3y =4
  #?

Savannah Anderson · Accepted Answer

The solution for the system of equations is:

#x=17/7#
#y=-2/7# #x-2y =3# , multiplying by #2# #color(blue)(2x)-4y =6#.......equation #(1)# #color(blue)(2x) +3y =4#........equation #(2)# Solving by elimination: subtracting equation #2# from #1#: #cancelcolor(blue)(2x)-4y =6# #-cancelcolor(blue)(2x) -3y =-4# #-7y = 2# #y=-2/7# Finding #x# from equation #1#: #x-2y =3# #x=3+2y# #x=3+2xx(-2/7)# #x=3- 4/7# #x=21/7- 4/7# #x=17/7#

Hazel Anglin · Answer

undefined To solve the system of equations:
$$ 
\begin{cases} 
x - 2y = 3 \
2x + 3y = 4 
\end{cases} 
$$

1. Multiply the first equation by 2 to eliminate x:
$$ 
\begin{cases} 
2x - 4y = 6 \
2x + 3y = 4 
\end{cases} 
$$

2. Subtract the second equation from the first equation:
$$ 
\begin{cases} 
(2x - 4y) - (2x + 3y) = 6 - 4 \
-7y = 2 \
\end{cases} 
$$

3. Solve for y:
$$ 
\begin{align*} 
-7y & = 2 \
y & = -\frac{2}{7} 
\end{align*} 
$$

4. Substitute the value of y into either of the original equations to solve for x. Using the first equation:
$$ 
\begin{align*} 
x - 2\left(-\frac{2}{7}\right) & = 3 \
x + \frac{4}{7} & = 3 \
x & = 3 - \frac{4}{7} \
x & = \frac{21}{7} - \frac{4}{7} \
x & = \frac{17}{7} 
\end{align*} 
$$

5. The solution to the system of equations is $ x = \frac{17}{7} $ and $ y = -\frac{2}{7} $.

How do you solve the following system: #x-2y=3, 2x+3y =4 #?