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2 years ago

−4+bx=2x+3(x+1) In the equation shown above, b is a constant. For what value of b does the equation have no solutions?

Mathematics

1 answer:

Likurg_2 [28]2 years ago

7 0

Answer:

b=5

Step-by-step explanation:

−4+bx=2x+3(x+1) (expand parenthesis first, i.e PEDMAS)

−4 + bx = 2x + 3x + 3

−4 + bx = 5x + 3 (move all x terms to one side and constants to another)

5x - bx = - 4 -3

5x - bx = - 7 (factor out x)

(5 - b)x = - 7

x = -7 / (5-b)

we can see that 5-b is in the denominator. Recall that for a fraction, the denominator cannot be zero otherwise the expression is undefined. (i.e no solutions)

hence the expression is undefined if,

(5-b) = 0

or b = 5

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2 years ago

What is the common difference of the sequence below?

vagabundo [1.1K]

Answer:

Step-by-step explanation:

In finding the COMMON DIFFERENCE, subtract the 2nd term and the first term.

a1 = -4

a2 = -2

Let "d" representing the COMMON DIFFERENCE.

d = -2 -(-4)

d = -2 + 4

d = 2

ANSWER:

THE COMMON DIFFERENCE OF THIS SEQUENCE IS 2

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2 years ago

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Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
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It is an extraneous solution.

$\therefore \: x = - 1$
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QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

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$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

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ivolga24 [154]

Translation to the left by 4 units

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Remember: Coplanar definition: In the same plane.

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3 years ago

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