3 years ago

Look at this system of equations. How can the system be changed so that the two equations have equal x-coefficients?

1 answer:

Volgvan3 years ago

5 0

I believe the correct answer from the choices listed above is option A. The <span>system can be changed so that the two equations have equal x-coefficients by multiplying </span><span>both sides of the top equation by 2 resulting to 6x + 4y = 24. Hope this answers the question.</span>

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Sonja [21]

Step-by-step explanation:

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3 years ago

Ignore circled one and work it out please

mash [69]

Answer:

a. $\frac{(x-7)}{x^2+x-30}$

Step-by-step explanation:

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