The equation of line k is y = mx + b, where m and b are some constants. What is the value of b?
(1) (b, 3b) belongs to k.
(2) (1, 1) belongs to k.

A. Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
B. Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
C. Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
D. Either statement BY ITSELF is sufficient to answer the question.
E. Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, meaning that further information would be needed to answer the question.

(E) Statement (1) tells us that point (b, 3b) belongs to line k. Let’s plug it into the equation of the line. 3b = mb + b. This yields b(m – 2) = 0. So either b = 0 or m = 2. We do NOT have a definite value of b if m = 2. Therefore statement (1) by itself is NOT sufficient.

Statement (2) tells us that (1, 1) belongs to k. Let’s plug it in the equation of the line. 1 = m × 1 + b. Clearly, not knowing the value of m, we can NOT find the value of b. Therefore statement (2) by itself is NOT sufficient.
When we use the both statements together, we have the following system of two equations:
b(m – 2) = 0
1 = m + b
If the solution of the first equation is b = 0, then the second one yields 1 = m + 0. So in this case m = 1.
If the solution of the first equation is m = 2, then the second one yields 1 = 2 + b. So in this case b = -1. Thus we have two different possible solutions of the system: b = 0, m = 1 (the equation of the line is y = x); b = -1, m = 2 (the equation of the line is y = 2x – 1). Two different values of b are possible. Therefore statements (1) and (2) taken together are NOT sufficient to answer the question. The correct answer is E.
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I did a mistake of cancelling out the b on both the sides for the statement 1 ... I don't understand that why we cant do it because it is only in the inequality that we should not cancel out the variables ...or does this rule apply to the equalities as well....

Please confirm

Division by 0 is impossible, so when you divide by a variable, you must be sure that it is NOT 0. Dividing the both sides of an equation by the same number, which is NOT 0, yields the equation with the same answers (it is a legitimate transformation).

In inequalities, you additionally must know the sign of the variable (is it a positive number, or a negative number?). Or you must consider two different cases:
- divide under the assumption that the variable is negative
- divide under the assumption that the variable is positive

We rearrange 3b = mb + b .

3b = mb + b
0 = mb + b – 3b
0 = mb – 2b
0 = b(m – 2)

This equality holds when m = 2 (b can be ANY)
0 = b × (2 – 2)
0 = b × 0

OR this equality holds when b = 0 (m can be ANY)
0 = 0 × (m – 2)

Thus we can NOT conclude that m is definitely 2.

If you divide by b you assume that it is not 0. In this case you solve not the original question, but the original question with the condition "b can not be equal to 0".

Therefore in order to solve the given question you must either not to divide by b, or consider two cases:
"b is NOT 0" (in this case you can divide by b)
and
"b = 0" (in this case you can use 0 instead of b) .