Can someone help me solve this math problem please?

Mathematics

1 answer:

SCORPION-xisa [38]3 years ago

7 0

2x-1=0
2x=1
x=1/2

x+5=0
x=-5

Final answer: B

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Solve the system of linear equations by substitution.<br><br> 8x−13y=0<br> 12x+3=y

Oliga [24]

Answer:

x = - 39/148

y = -6/37

Step-by-step explanation:

substitute the given value of y into the equation to get 8x-13 (12x+3)=0

solve for x to get x= - 39/148

substitute the given value of x into 12x+3=y

12 x (-39/148) + 3 = y

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7 0

2 years ago

What is lim x→3 x^2 - x - 6/ x - 3

Kisachek [45]

<h3>Answer: D) 5</h3>

=======================================================

Explanation:

If we plugged x = 3 into the expression, then we'd get x-3 = 3-3 = 0 in the denominator. That's not allowed. But we can simplify first

x^2-x-6 factors to (x-3)(x+2). The key here is that (x-3) is a factor. It cancels with the x-3 in the denominator

So, $\frac{x^2-x-6}{x-3} = \frac{(x-3)(x+2)}{x-3} = x+2$

Allowing us to say,

$\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}\frac{(x-3)(x+2)}{x-3}\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = \lim_{x \to 3}(x+2)\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 3+2\\\\\\\displaystyle \lim_{x \to 3}\frac{x^2-x-6}{x-3} = 5\\\\\\$