How do you solve the following system: #4x+3y= -7, y + 4x = 16 #?

Question

Eli Assouline · Accepted Answer

The solution for the system of equations is:
#color(blue)(x=55/8#

#color(blue)(y=-23/2# #color(blue)(4x)+3y=-7#..........equation #1# #color(blue)(4x)+y=16#.................equation #2# Solving by elimination Subtracting equation #2# from #1# #cancelcolor(blue)(4x)+3y=-7# #-cancelcolor(blue)(4x)-y=-16# #2y=-23# #color(blue)(y=-23/2# Finding #x# from equation #1# #4x+3y=-7# #4x+3xx(-23/2)=-7# #4x-69/2=-7# #4x=-7 +69/2# #4x=-14/2 +69/2# #4x=55/2# #x=55/(2xx4)# #color(blue)(x=55/8#

Hazel Anglin · Answer

undefined To solve the system of equations $4x + 3y = -7$ and $y + 4x = 16$, you can use the method of substitution or elimination. Let's use the substitution method.

First, solve one of the equations for one variable. Let's solve the second equation for $y$:

$$y = 16 - 4x$$

Now, substitute this expression for $y$ into the first equation:

$$4x + 3(16 - 4x) = -7$$

Now, solve for $x$:

$$4x + 48 - 12x = -7$$
$$-8x + 48 = -7$$
$$-8x = -55$$
$$x = \frac{-55}{-8} = \frac{55}{8}$$

Now that we have found $x$, substitute this value back into either of the original equations to find $y$. Let's use the second equation:

$$y + 4\left(\frac{55}{8}\right) = 16$$
$$y + \frac{55}{2} = 16$$
$$y = 16 - \frac{55}{2}$$
$$y = \frac{32}{2} - \frac{55}{2}$$
$$y = \frac{-23}{2}$$

So, the solution to the system of equations is $x = \frac{55}{8}$ and $y = \frac{-23}{2}$.