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2 years ago

Py+7=6y+q what is the answer for this question (linear equations with unknown coefficients)?

Mathematics

2 answers:

Temka [501]2 years ago

3 0

Assuming your solving for y:
py+7=6y+q
-6y -7 -6y -7
(p-6)y = q-7
divide both sides by p-6
y=(q-7)/(p-6)

kompoz [17]2 years ago

3 0

Answer:

The required equation is $y =\frac{(q-7)}{(p-6)}$ .

Step-by-step explanation:

Consider the provided equation.

$py+7=6y+q$

We need to simplify the equation and convert it into a linear equation with unknown coefficients.

Subtract 6y from both sides.

$py - 6y + 7 = q$

Subtract 7 from both sides.

$py - 6y = q - 7$

Take y common from left side.

$y(p-6) = q-7$

Divide both sides by p-6.

$y =\frac{(q-7)}{(p-6)}$

Hence, the required equation is $y =\frac{(q-7)}{(p-6)}$ .

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ziro4ka [17]

Answer: true

Step-by-step explanation:

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where a1, a2, ... are scalars, and c1 is the vector of column 1.

Then we should write:

Column space = a1*(A₁₁, A₂₁, A₃₁, ...) + a2*(A₁₂, A₂₂, A₃₂, ...) + ...

Now, the transpose is defined as:

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Here i used the element with subindex 3 and 1, but is the same for every subindex.

Notice that if A is m*n, then [At] is n*m

Now, the row space of [At] will be, same as before.

Row space = b1*r1 + b2*r2 + ...

Where b1, b2, ... are scalars and the r's are the vector of each row.

= b1*( [At]₁₁ , [At]₁₂, [At]₁₃, ...) + b2*([At]₂₁, [At]₂₂, [At]₂₃, ...) + ...

Now we replace each term of the transpose by the associated element in the original matrix.

= b1*( A₁₁, A₂₁, A₃₁, ...) + b2*(A₂₁, A₂₂, ...) + ....

If we take:

b1 = a1, b2 = a2, b3 = a3, ...

We will have that the row space of [At] is the same as the column space of A.

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