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Sagot :
Considering the form:
[tex]y=mx+b[/tex]where y is the variable that depends on x, m changes depending on the value of x, and b is constant.
Then, in our problem:
• y ,is the cost of the chauffeured car service for both equations.
,• x ,is the miles traveled.
,• m ,is 4 in the first and 2 in the second.
,• b ,is 3 in the first and 19 in the second.
Then, replacing the latter we can build our equations.
• Equation 1 ,(first service)
[tex]y=4x+3[/tex]• Equation 2 ,(second service)
[tex]y=2x+19[/tex]Solving the system of equations by substitution.
0. Replacing ,Equation 2 ,in equation ,1,:
[tex]y=4x+3[/tex][tex]2x+19=4x+3[/tex]2. Grouping similar terms and solving for x
[tex]19-3=4x-2x[/tex][tex]16=2x[/tex][tex]x=\frac{16}{2}[/tex][tex]x=8[/tex]3. Replacing this value in Equation 2 to get y:
[tex]y=2\cdot8+19[/tex][tex]y=16+19[/tex][tex]y=35[/tex]Answer: at 5 miles the two companies charge the same total fare of $35.
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