How do i solve equations in the form px+q=r and p(x+q) = r

Explanation:

In general, you solve equations by "undoing" what was "done" to the variable, generally in the reverse of the order in which the operations were done.

These are known as "2-step" linear equations, because they can be solved in two steps.

In the first equation, the variable is multplied by p and q is added to the product. Reversing these steps in reverse order, we subtract q, then divide by p:

px +q = r . . . . . given equation

px = r - q . . . . . q is subtracted (from both sides)

x = (r -q)/p . . . . both sides are divided by p

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The rules of equality require that the same operation be performed on both sides of the equation. When we say "subtract q" we mean "subtract q from both sides of the equation." Doing something different to one side or the other will make it be a different equation than the one you're trying to solve. My teacher would say, "keep the equal sign sacred."

In the second equation, q is added to the variable, then the sum is multiplied by p. Reversing these steps in reverse order means we divide by p, then subtract q.

p(x +q) = r . . . . . given equation

x + q = r/p . . . . . divide by p

x = (r/p) -q . . . . . subtract q (from the quotient)

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Please note that the second equation can be made to look like the first equation if parentheses are eliminated.

p(x +q) = r

px +pq = r . . . . . . a constant added to the product of p and x

Using the first solution with these values, we find ...

x = (r -pq)/p

If we divide this out, we see ...

x = (r/p) - q . . . . . same as our solution to the second equation

(1, –1), (16, –4), (36, –6)

step-by-step explanation:

given that the function is

this can also be written as

we have to graph this function and check which list of points lie on the graph

see attachment for graph:

we find that the points lying are (1,-1) (16,-4) and (36,-6)

since square root is there we find that x can only be posiitve but y can be since square root both positive and negative.