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3 years ago

I need help with part B.

Mathematics

1 answer:

Montano1993 [528]3 years ago

6 0

Answer is 8/10 because the original fraction was 4/5 and they had 10 pizzas and that would be the denominator so you have to get the denominator to be 10 so you would multiply 4/5×2 and you would get 8/10 and since the denominator is 10 that is your answer think of it like this... part at the top whole at the bottom hope this helps

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Factor out the coefficient of the variable 1.7j-3.4

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Answer:

$\boxed {x_{1} = 3}$

$\boxed {x_{2} = -4}$

Step-by-step explanation:

Solve the following equation:

$\left[\begin{array}{ccc}3&2\\5&5\\\end{array}\right] \left[\begin{array}{ccc}x_{1}\\x_{2}\\\end{array}\right] + \left[\begin{array}{ccc}1\\2\\\end{array}\right] = \left[\begin{array}{ccc}2\\-3\\\end{array}\right]$

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$3x_{1} + 2x_{2} + 1 - 1 = 2 - 1$

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$3x_{1} = -2x_{2} + 1$

-Divide both sides by $3$ :

$3x_{1} = -2x_{2} + 1$

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

-Multiply $-2x_{2} + 1$ by $\frac{1}{3}$ :

$x_{1} = \frac{1}{3} (-2x_{2} + 1)$

$x_{1} = -\frac{2}{3}x_{2} + \frac{1}{3}$

-Substitute $-\frac{2x_{2} + 1}{3}$ for $x_{1}$ in the second equation, which is $5x_{1} + 5x_{2} + 2 = -3$ :

$5x_{1} + 5x_{2} + 2 = -3$

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

Multiply $-\frac{2x_{2} + 1}{3}$ by $5$ :

$5(-\frac{2}{3}x_{2} + \frac{1}{3}) + 5x_{2} + 2 = -3$

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$\frac{5}{3}x_{2} = -\frac{20}{3}$

-Multiply both sides by $\frac{5}{3}$ :

$\frac{\frac{5}{3}x_{2}}{\frac{5}{3}} = \frac{-\frac{20}{3}}{\frac{5}{3}}$

$\boxed {x_{2} = -4}$

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$x_{1} = \frac{9}{3}$

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Answer:

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