Given the system of equations:
[tex]\begin{gathered} 2x+9y=27 \\ x-3y=-24 \end{gathered}[/tex]
To solve it by substitution, follow the steps below.
Step 1: Solve one linear equation for x in terms of y.
Let's choose the second equation. To solve it for x, add 3y to each side of the equations.
[tex]\begin{gathered} x-3y=-24 \\ x-3y+3y=-24+3y \\ x=-24+3y \end{gathered}[/tex]
Step 2: Substitute the expression found for x in the first equation.
[tex]\begin{gathered} 2x+9y=27 \\ 2\cdot(-24+3y)+9y=27 \\ -48+6y+9y=27 \\ -48+15y=27 \end{gathered}[/tex]
Step 3: Isolate y in the equation found in step 2.
To do it, first, add 48 to both sides.
[tex]\begin{gathered} -48+15y=27 \\ -48+15y+48=27+48 \\ 15y=75 \end{gathered}[/tex]
Then, divide both sides by 15.
[tex]\begin{gathered} \frac{15y}{15}=\frac{75}{15} \\ y=5 \end{gathered}[/tex]
Step 4: Substitute y by 5 in the relation found in step 1 to find x.
[tex]\begin{gathered} x=-24+3y \\ x=-24+3\cdot5 \\ x=-24+15 \\ x=-9 \end{gathered}[/tex]
Answer:
x = -9
y = 5
or (-9, 5)
Also, you can graph the lines by choosing two points from each equation, according to the picture below.
