Solve the following equation for x 10=2-4(ax-3)

Mathematics

1 answer:

likoan [24]3 years ago

7 0

Answer:

$\large\boxed{x=\dfrac{1}{a}}$

Step-by-step explanation:

$2-4(ax-3)=10\qquad\text{subtract 2 from both sides}\\\\-4(ax-3)=8\qquad\text{divide both sides by (-4)}\\\\\dfrac{-4(ax-3)}{-4}=\dfrac{8}{-4}\\\\ax-3=-2\qquad\text{add 3 to both sides}\\\\ax-3+3=-2+3\\\\ax=1\qquad\text{divide both sides by}\ a\neq0\\\\x=\dfrac{1}{a}$

You might be interested in

I need help with this math question :(

Ray Of Light [21]

Answer:

k = 5,

y = 5x

Step-by-step explanation:

If y is proportional to x,

y ∝ x

Then the equation will be

y = kx

Her k = proportionality constant.

Now we choose a point from the given table.

Let the point is (2, 10).

By putting the values of x and y in the equation,

10 = 2×k

k = $\frac{10}{2}$

k = 5

This will be true for any point we choose from the table.

And the equation will be y = 5x

8 0

3 years ago

What is the row echelon form of this matrix?

USPshnik [31]

The row echelon form of the matrix is presented as follows;

$\begin{bmatrix}1 &-2 &-5 \\ 0& 1 & -7\\ 0&0 &1 \\\end{bmatrix}$

<h3>What is the row echelon form of a matrix?</h3>

The row echelon form of a matrix has the rows consisting entirely of zeros at the bottom, and the first entry of a row that is not entirely zero is a one.

The given matrix is presented as follows;

$\begin{bmatrix}-3 &6 &15 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}$

The conditions of a matrix in the row echelon form are as follows;

There are row having nonzero entries above the zero rows.
The first nonzero entry in a nonzero row is a one.
The location of the leading one in a nonzero row is to the left of the leading one in the next lower rows.

Dividing Row 1 by -3 gives:

$\begin{bmatrix}1 &-2 &-5 \\ 2& -6 & 4\\ 1&0 &-1 \\\end{bmatrix}$