If x and y in the equations

1 answer:

5 0

We can let
a = 1/(x-1)
b = 1/(y+2)
and rewrite the equations as
2a - b = 10
a + 3b = -9
Using the first to write an expression for b, we get
b = 2a - 10
Substituting this into the second equation gives
a + 3(2a -10) = -9
7a -30 = -9 . . . . . . . . simplify
7a = 21 . . . . . . . . . . .add 30
a = 3
b = 2·3 - 10 = -4

Now, we can find x and y.
   3 = 1/(x -1)
   x - 1 = 1/3
   x = 1 1/3 = 4/3

  -4 = 1/(y +2)
  y +2 = -1/4
  y = -2 1/4 = -9/4

Then the desired sum is
  x + y = 4/3 -9/4 = (16 -27)/12
x + y = -11/12

The appropriate choice is ..
c. -11/12

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The answer to this question is:

C) 8.

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3 years ago

Add. Show Your Work.

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Hey there :)

$1 \frac{4}{5} + 1\frac{6}{10}$

Change mixed fraction into an improper fraction
$\frac{9}{5} + \frac{16}{10}$
$\frac{16}{10}$ : are both even numbers, so you can further simplify by dividing by 2
$\frac{16/2}{10/2} = \frac{8}{5}$

To add fractions, we need same denominator to proceed. Here we have no problem since both denominators are 5

$\frac{9}{5}+ \frac{8}{5} = \frac{9+8}{5}$
$\frac{17}{5}$

As a mixed fraction: $3 \frac{2}{5}$
______________________________________________________

$2 \frac{3}{4} - 1\frac{2}{8}$

Let us change the mixed fraction into an improper fraction:
$\frac{11}{4} - \frac{10}{8}$
$\frac{10}{8}$ : You can further simplify this since both are divisible by 2
$\frac{10/2}{8/2} = \frac{5}{4}$

We have the same denominator set for subtraction, so you do not have to find common denominators
$\frac{11}{4}- \frac{5}{4} = \frac{11-5}{4}$
$\frac{6}{4}$

As a mixed fraction: $1\frac{2}{4} = 1\frac{1}{2}$

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3 years ago

Giving brainliest and max rating.

Alexus [3.1K]

Answer:

5+-2

Step-by-step explanation:

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3 years ago

Read 2 more answers

130 students choose to attend one of three after school activities: football, tennis or running. There are 58 boys. 49 students

cestrela7 [59]

Answer: $\dfrac{27}{65}$

Step-by-step explanation:

There are 130 students.

There are 58 boys --> 72 girls

A) 49 chose football: 27 are girls --> 22 are boys

B) 72 girls: 24 chose running, 27 chose football --> 21 girls chose tennis

C) 27 students chose tennis: 21 are girls --> 6 are boys.

D) 58 boys: 22 chose football, 6 chose tennis --> 30 boys chose running.

$\large\boxed{\begin{array}{l|cc||c}&\underline{Boys}&\underline{Girls}&\underline{Total}\\Football&22&27&49\\Tennis&6&21&27\\\underline{Running}&\underline{\quad 30\quad}&\underline{\quad 24\quad}&\underline{\quad 54\quad}\\Total&58&72&130\end{array}}$