How do you solve the system #x = y + 4# and #2x - 5y = 2# by substitution?

Question

Isaiah Anthony · Accepted Answer

#x = y + 4#————(1) #2x - 5y = 2#————— (2)

The following results from entering #x# into the second equation:

#2(y+4) - 5y = 2#

$2y+8 - 5y = 2#

# -3y + 8 = 2#

#-3y = 2–8#

#-3y = -6#

When we divide both sides by #-3#, we obtain:

(-6)/-3# = #(cancel(-3)y)/cancel(-3)

#color(y = 2#) for green

The following results from changing #y=2# in the first equation:

#x equals 2 + 4#

Green #color (x = 6)

The answers to the following two equations are: #x = 6; y = 2#

Confirm:

Replace #x and y# with their respective values in each equation to check if they are satisfied.

Jace Anderson · Answer

undefined To solve the system $x = y + 4$ and $2x - 5y = 2$ by substitution:

1. Solve the first equation for one variable (let's solve for $x$): $x = y + 4$.
2. Substitute the expression for $x$ from the first equation into the second equation.
3. Solve the resulting equation for the variable left.

Here's the process:

$2x - 5y = 2$ (Given second equation)

Substitute $x = y + 4$ into the second equation:

$2(y + 4) - 5y = 2$

Now, solve for $y$:

$2y + 8 - 5y = 2$

$-3y + 8 = 2$

$-3y = -6$

$y = 2$

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

$x = y + 4$

$x = 2 + 4$

$x = 6$

So, the solution to the system is $x = 6$ and $y = 2$.