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2 years ago

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In your notebook, set up the following subtraction in a vertical format and select the correct answer. Find 2p 2 3p - 4 less - 2

p 2 - 3p 4. 4p2 6p 8 4p2 - 6p - 8 4p2 6p - 8 0.

1 answer:

Elena L [17]2 years ago

3 0

The equation in the reduced form of the equation is $\rm 4p^ 2 + 6p - 8$ .

Given that,

Equation; $\rm 2p^ 2 + 3p - 4\ by\ -2p ^2 - 3p +4$

We have to reduce the equation.

According to the question,

To reduce the equation means we need to subtract one equation from the other.

To determine the reduced form of the equation following all the steps given below.

Equation; $\rm 2p^ 2 + 3p - 4\ by\ -2p ^2 - 3p +4$

Subtraction equation 1 from equation 2,

$\rm = \rm 2p^ 2 + 3p - 4\ -(-2p ^2 - 3p + 4)\\\\= 2p^ 2 + 3p - 4\ -(-2p ^2 -3p +4)\\\\ = 2p^ 2 + 3p - 4+2p^ 2 + 3p - 4 \\\\ = 4p^ 2 + 6p - 8$

Hence, The equation in the reduced form of the equation is $\rm 4p^ 2 + 6p - 8$ .

For more details refer to the link given below.

brainly.com/question/1807316

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Infinite number of solutions.

Step-by-step explanation:

We are given system of equations

$5x+4y+5z=-1$

$x+y+2z=1$

$2x+y-z=-3$

Firs we find determinant of system of equations

Let a matrix A= $\left[\begin{array}{ccc}5&4&5\\1&1&2\\2&1&-1\end{array}\right]$ and B= $\left[\begin{array}{ccc}-1\\1\\-3\end{array}\right]$

$\mid A\mid=\begin{vmatrix}5&4&5\\1&1&2\\2&1&-1\end{vmatrix}$

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Determinant of given system of equation is zero therefore, the general solution of system of equation is many solution or no solution.

We are finding rank of matrix

Apply $R_1\rightarrow R_1-4R_2$ and $R_3\rightarrow R_3-2R_2$

$\left[\begin{array}{ccc}1&0&1\\1&1&2\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\1\\-5\end{array}\right]$

Apply $R_2\rightarrow R_2-R_1$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&-1&-3\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\-5\end{array}\right]$

Apply $R_3\rightarrow R_3+R_2$

$\left[\begin{array}{ccc}1&0&1\\0&1&1\\0&0&-2\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\6\\1\end{array}\right]$

Apply $R_3\rightarrow- \frac{1}{2}$ and $R_2\rightarrow R_2-R_3$

$\left[\begin{array}{ccc}1&0&1\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-5\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Apply $R_1\rightarrow R_1-R_3$

$\left[\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right]$ : $\left[\begin{array}{ccc}-\frac{9}{2}\\\frac{13}{2}\\-\frac{1}{2}\end{array}\right]$

Rank of matrix A and B are equal.Therefore, matrix A has infinite number of solutions.

Therefore, rank of matrix is equal to rank of B.

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