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Subtracting the second equation from the first eliminates b and gives 30a= 50 so a= 5/3. Then 100a+ b= 500/3+ b= 100 so b= 100- 500/3= -200/3. f(x)= (5/3)x- 200/3= (5x- 200)/3.

Checking, f(100)= (500- 200)/3= 300/3= 100 and f(70)= (350- 200)/3= 150/3= 50.

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It would be better if you state which part(s) of the explanation you don't understand. You are given two points on the graph of the function:Hi hallsoflvy, could you elaborate a bit im lost.

Thank

(70,50) and (100,100)

As HallsofIvy pointed out, there are an infinite number of functions we could find passing through those two points, but the simplest of these is going to be a linear function (\(n\) points can be uniquely described by a polynomial of degree \(n-1\)). Using the definition of slope, and the point-slope equation for a line, we then find:

\(\displaystyle f(x)=\frac{100-50}{100-70}(x-70)+50=\frac{5}{3}x-70\cdot\frac{5}{3}+50=\frac{5}{3}x-\frac{200}{3}=\frac{5}{3}(x-40)\)

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We could, if you could explain where you're stuck.… could you elaborate …

If you're not familiar with linear equations (eg: lines, slopes, equation forms), let us know. We can help you find online videos and lesson links.

Please read the forum's submission guidelines. You can start with

It would be better if you state which part(s) of the explanation you don't understand. You are given two points on the graph of the function:

(70,50) and (100,100)

As HallsofIvy pointed out, there are an infinite number of functions we could find passing through those two points, but the simplest of these is going to be a linear function (\(n\) points can be uniquely described by a polynomial of degree \(n-1\)). Using the definition of slope, and the point-slope equation for a line, we then find:

\(\displaystyle f(x)=\frac{100-50}{100-70}(x-70)+50=\frac{5}{3}x-70\cdot\frac{5}{3}+50=\frac{5}{3}x-\frac{200}{3}=\frac{5}{3}(x-40)\)

Thanks, this cleared my confusion