Answer:
[tex](-2, -4)[/tex]
Step-by-step explanation:
To solve the system of equations [tex]x = 3y + 10[/tex] and [tex]x = y + 2[/tex] by substitution, we'll substitute the expression for [tex]x[/tex] from the second equation into the first equation.
Given:
[tex] \begin{cases} x = 3y + 10 \textsf{....... Equation 1} \\\ x = y + 2\textsf{......... Equation 2} \end{cases} [/tex]
We'll substitute the expression for [tex]x[/tex] from the second equation into the first equation:
[tex] y + 2 = 3y + 10 [/tex]
Now, we'll solve for [tex]y[/tex]:
[tex] y + 2 -2 -3y = 3y + 10-2-3y [/tex]
[tex] y - 3y = 10 - 2 [/tex]
[tex] -2y = 8 [/tex]
[tex] \dfrac{-2y }{-2}=\dfrac{ 8 }{-2} [/tex]
[tex] y = -4 [/tex]
Now that we've found [tex]y = -4[/tex], we can substitute this value back into one of the original equations to find [tex]x[/tex].
Let's use the second equation:
[tex] x = (-4) + 2 [/tex]
[tex] x = -2 [/tex]
So, the ordered pair that satisfies both equations is [tex](-2, -4)[/tex].
