At what point does the graph of 5x + 4y = 12 intersect the y-axis?

1 answer:

4 0

To find the y-intercept, we need to re-arrange the equation given to y=mx+c, where c is the y-intercept
To do this to 5x + 4y = 12, we need to have only y on the left side, so subtract 5x from each side, to get
4y = 12 - 5x, or
4y = -5x + 12
Now we need to isolate y, so divide each side by 4
y = 5/4x + 3
So the y-intercept will be at +3, and the y intercept is always on x = 0, so the co-ordinate will be (0, 3), which is a.

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Recall that a relation is an equivalence relation if and only if is symmetric, reflexive and transitive. In order to simplify the notation we will use A↔B when A is in relation with B.

Reflexive: We need to prove that A↔A. Let us write J for the identity matrix and recall that J is invertible. Notice that $A=J^{-1}AJ$ . Thus, A↔A.

Symmetric: We need to prove that A↔B implies B↔A. As A↔B there exists an invertible matrix P such that $A=P^{-1}BP$ . In this equality we can perform a right multiplication by $P^{-1}$ and obtain $AP^{-1} =P^{-1}B$ . Then, in the obtained equality we perform a left multiplication by P and get $PAP^{-1} =B$ . If we write $Q=P^{-1}$ and $Q^{-1} = P$ we have $B = Q^{-1}AQ$ . Thus, B↔A.

Transitive: We need to prove that A↔B and B↔C implies A↔C. From the fact A↔B we have $A=P^{-1}BP$ and from B↔C we have $B=Q^{-1}CQ$ . Now, if we substitute the last equality into the first one we get