What is the ordered pair for y=4x+6 when x=−3?

Ivan Buckley
2022-09-23
Answered

What is the ordered pair for y=4x+6 when x=−3?

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Miya Swanson

Answered 2022-09-24
Author has **11** answers

It is pretty simple when you think about it. They give you the x value, so all you do is sub it into the equation y=4x+6.

y=4(−3)+6

4 times -3 will give us -12

y=−12+6

−12+6 gives us -6

So our final answer ends up being

y=−6

y=4(−3)+6

4 times -3 will give us -12

y=−12+6

−12+6 gives us -6

So our final answer ends up being

y=−6

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We know the standard form of expressing a system of linear equations in n variables in n equations.

$A}_{n\times n}\cdot {X}_{n\times 1}={B}_{n\times 1$

Where A is the coefficient matrix, X is the unknown variables (each variable maybe a real or complex number) matrix and B is the constants matrix.

Now, arriving at the question, let's say I have a set of n vectors X, each of length n and another set of n+1 vectors Y, each of length n. (Clearly,$size\left(X\right)<size\left(Y\right)$ ). All elements of X are independent. The same goes for Y.

Now I want to express each element in Y, as a linear expression of all the elements in X. So that would be something like below, for some coefficient matrix K.

$K}_{(n+1)\times n}\cdot {X}_{n\times n}={Y}_{(n+1)\times n$

My question is, is it necessary that K must always exist i.e. there must be a way to express all elements in Y as a linear combination of X.

I think the answer is No. The reasoning is that, for a system of linear equations where number of equations is strictly greater than the number of unknowns, there does not exist a solution at all. In the above case, we have n unknowns from X and n+1 equations, each one for each element in Y.

Is my answer and reasoning correct?

Where A is the coefficient matrix, X is the unknown variables (each variable maybe a real or complex number) matrix and B is the constants matrix.

Now, arriving at the question, let's say I have a set of n vectors X, each of length n and another set of n+1 vectors Y, each of length n. (Clearly,

Now I want to express each element in Y, as a linear expression of all the elements in X. So that would be something like below, for some coefficient matrix K.

My question is, is it necessary that K must always exist i.e. there must be a way to express all elements in Y as a linear combination of X.

I think the answer is No. The reasoning is that, for a system of linear equations where number of equations is strictly greater than the number of unknowns, there does not exist a solution at all. In the above case, we have n unknowns from X and n+1 equations, each one for each element in Y.

Is my answer and reasoning correct?