The form of the linear equation is
[tex]y=mx+b[/tex]m is the slope
b is the y-intercept (initial value at x = 0)
Since the cost is C and the number of dances is L, then
The equation should be
[tex]C=mL+b[/tex]We will use the given information to find m and b
Since 7 lessons cost $82 dollars, then
Substitute C by 82 and L by 7
[tex]\begin{gathered} 82=m(7)+b \\ 82=7m+b \\ 7m+b=82\rightarrow(1) \end{gathered}[/tex]Since 11 lessons cost $122, then
Substitute C by 122 and L by 11
[tex]\begin{gathered} 122=m(11)+b \\ 122=11m+b \\ 11m+b=122\rightarrow(2) \end{gathered}[/tex]Now, we have a system of equations to solve it
Subtract equation (1) from equation (2) to eliminate b
[tex]\begin{gathered} (11m-7m)+(b-b)=(122-82) \\ 4m+0=40 \\ 4m=40 \end{gathered}[/tex]Divide both sides by 4 to find m
[tex]\begin{gathered} \frac{4m}{4}=\frac{40}{4} \\ m=10 \end{gathered}[/tex]Substitute m in equation (1) by 10 to find b
[tex]\begin{gathered} 7(10)+b=82 \\ 70+b=82 \end{gathered}[/tex]Subtract 70 from both sides to find b
[tex]\begin{gathered} 70-70+b=82-70 \\ b=12 \end{gathered}[/tex]The equation of the cost of L lessons is
[tex]C=10L+12[/tex]If the number of the lessons is 4, then substitute L by 4 to find the cost
[tex]\begin{gathered} C=10(4)+12 \\ C=40+12 \\ C=52 \end{gathered}[/tex]The cost of the 4 lessons is $52
11. C = 10L + 12
12. $52
