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3 years ago

4x-8y+16x+9y-5x²-8x+6x²

Mathematics

2 answers:

olga2289 [7]3 years ago

3 0

4x-8y+16x+9y-5x^2-8x+6x^2=

12x+1y+1x^2=

x^2+12x+1y

NISA [10]3 years ago

3 0

1) group the same variables together
6x²-5x²+16x-8x+9y-8y
TIP: PARENTHESIS MIGHT HELP
(6x²-5x²)+(16x-8x)+(9y-8y)

2) solve each group
6x²-5x²= 1x² AKA x²
16x-8x= 8x
9y-8y= 1y AKA y

3) put the equation back together
x²+8x+y

THE FINAL ANSWER IS…
x²+8x+y

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Read 2 more answers

The matrix below represents a system of equations.

noname [10]

Answer:

The first choice: $a = 1$ , $b = -1$ , and $c = 2$ .

Step-by-step explanation:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 1 & -1 & -2 \cr 1 & 0 & -3 & -5\end{array} \right]$ .

Add the first row to the second row to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 2 + 1 & (-1) + 1 & 1 + (-1) & 5 + (-2) \cr 1 & 0 & - 3& -5\end{array} \right]$ .

Simplify that matrix to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 3 & 0 & 0 & 3\cr 1 & 0 & -3 & -5\end{array} \right]$ .

Divide the second row by $3$ :

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 1 & 0 & -3& -5\end{array} \right]$ .

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Subtract the current row two from row three:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 1 - 1 & 0 & -3 & (-5)- 1 \end{array} \right]$ .

Simplify that matrix to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 0 & 0 & -3 & -6 \end{array} \right]$ .

Invert the signs in the third row and divide it by $3$ to obtain:

$\left[ \begin{array}{ccc|c} 2 & -1 & 1 & 5 \cr 1 & 0 & 0 & 1\cr 0 & 0 & 1 & 2 \end{array} \right]$ .

Hence, $c = 2$ .

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That simplifies to

$\left[ \begin{array}{ccc|c} 0 & -1 & 1 & 3\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$ .

Subtract the current third row from the first row to obtain:

$\left[ \begin{array}{ccc|c} 0 & -1 & 1-1 & 3-2\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$ .

That simplifies to

$\left[ \begin{array}{ccc|c} 0 & -1 & 0& 1\cr 1 & 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$ .

Invert the signs in the first row to obtain:

$\left[ \begin{array}{ccc|c} 0 & 1 & 0& -1\cr 1& 0 & 0 & 1\cr 0 & 0 & 1& 2\end{array} \right]$ .

Hence, $b = -1$ .

$\left[ \begin{array}{ccc|c}1& 0 & 0 & 1 \cr0 & 1 & 0& -1 \cr 0 & 0 & 1& 2\end{array} \right]$ .

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