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3 years ago

3x-6y=9 x=2y+3 SOMEONE HELP ME SOLVE THIS STEP BY STEP TY

Mathematics

1 answer:

RideAnS [48]3 years ago

6 0

Answer:

0=0

infinitely many solutions

Step-by-step explanation:

plug x=2y+3 as x into 3x-6y=9...so

3(2y+3)-6y=9, which has our x=2y+3 plugged in because that is what x equals

if you solve its

6y+9-6y=9

0y=0

y=0

making it 0=0

you can't go further from this

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Answer:

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Step-by-step explanation:

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Combine the bottom, too.

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Apply the fraction rule

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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

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Answer:

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Step-by-step explanation:

$\huge \frac{x^2-3x-28}{(x^2+10x+24)(x-5)} \\\\\huge {=\frac{x^2-7x+4x-28}{(x^2+6x+4x+24)(x-5)}}\\\\\huge {=\frac{x(x-7)+4(x-7)}{\{x(x+6)+4(x+6)\}(x-5)}}\\\\\huge {=\frac{(x-7)(x+4)}{(x+6)(x+4)(x-5)}}\\\\\huge {=\frac{(x-7)}{(x+6)(x-5)}}\\\\\huge {=\frac{(x-7)}{x^2+(6-5)x+6(-5)}}\\\\\huge {=\frac{(x-7)}{x^2+x-30}}$

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