+318 Two students in Mr. Delgado's class, Cole and Terrence, have been assigned a workbook to complete at their own pace. They got together at Cole's house after school to complete as many pages as they can. Cole has already completed 51 pages and will continue working at a rate of 5 pages per hour. Terrence has completed 41 pages and can work at a rate of 15 pages per hour. Eventually, the two students will be working on the same page. How long will that take? How many pages will each of them have completed?
+318 Before you can solve, you must write simultaneous equations. Let x represent hours, and let y represent pages.
y = 5x + 51
y = 15x + 41
Now use substitution to solve the simultaneous equations.
Step 1: Isolate a variable.
The variable y is already isolated in the first equation.
Step 2: Plug the result of Step 1 into the other equation and solve for one variable.
Plug y = 5x + 51 into the other equation, y = 15x + 41, and find the value of x.
y = 15x + 41
5x + 51 = 15x + 41
Plug in y = 5x + 51
–10x + 51
= 41 Subtract 15x from both sides
–10x = –10 Subtract 51 from both sides
x = 1 Divide both sides by –10
Step 3: Plug the result of Step 2 into one of the original equations and solve for the other variable.
Take the result of Step 2, x = 1, and plug it into one of the original equations, such as y = 5x + 51. Then find the value of y.
y = 5x + 51
y = 5(1) + 51 Plug in x = 1
y = 5 + 51 Multiply
y = 56 Add
Step 4: State the solution.
Since x = 1 and y = 56, the solution is (1, 56).
After 1 hour, Cole and Terrence will have each completed 56 pages in their workbooks.
+36916 c = 51 + 5x
t = 41 + 15x equate them to find the time , x when they are on the same page
51 + 5x = 41+ 15x
x = 1 hrs
cole will have 51 + 5(1) = 56 pages ..... and so will terrence
