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3 years ago

(2x + 3y) + (4x - 3y) = 11+(-5)

Mathematics

1 answer:

Maurinko [17]3 years ago

3 0

Answer:

The correct option should have been $6x=6$ .

Step-by-step explanation:

Given the expression

$\left(2x\:+\:3y\right)\:+\:\left(4x\:-\:3y\right)\:=\:11+\left(-5\right)$

solving the expression

$\left(2x\:+\:3y\right)\:+\:\left(4x\:-\:3y\right)\:=\:11+\left(-5\right)$

Remove parentheses: (a) = a

$2x+3y+4x-3y=11-5$

Group like terms

$3y-3y+2x+4x=11-5$

Add similar elements

$6x=6$

It is clear that not a single given option is $6x=6$ . It means no option is correct. It seems you mistyped the correct options.

The correct option should have been $6x=6$ .

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Answer:

3/2

Step-by-step explanation:

2y+3x=3

This represents a linear equation and the format for a linear equation is

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and divide both sides by 2

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3 years ago

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Step-by-step explanation:

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3 years ago

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2 years ago

What are the zeros of the function f(x) = (3x^2 + 9x + 6) / (3x - 6) ?

cricket20 [7]

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3x(x + 1) + 6(x + 1) = 0
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3 years ago

Simplify . (1/c + 1/h)/(1/(c ^ 2) - 1/(r ^ 2))

Marrrta [24]

Answer:

$\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Step-by-step explanation:

$\frac{\frac{1}{c}+\frac{1}{h}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine $\frac{1}{c} + \frac{1}{h}$

$\frac{\frac{h+c}{ch}}{\frac{1}{c^2}-\frac{1}{r^2}}$

Combine the bottom, too.

$=\frac{\frac{h+c}{ch}}{\frac{r^2-c^2}{c^2r^2}}$

Apply the fraction rule

$=\frac{\left(h+c\right)c^2r^2}{ch\left(r^2-c^2\right)}$

Cancel

$=\frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

Therefore, $\frac{\left(\frac{1}{c}+\frac{1}{h}\right)}{\left(\frac{1}{\left(c^2\right)}-\frac{1}{\left(r^2\right)}\right)}:\quad \frac{\left(h+c\right)cr^2}{h\left(r^2-c^2\right)}$

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3 years ago