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tigry1 [53]
3 years ago
11

6j-k=233k+6j=11Solve j and k ​using the elimination method on the simultaneous equations

Mathematics
1 answer:
frez [133]3 years ago
5 0

Answer:

Step-by-step explanation:

6j - k = 23

6j + 3k = 11.....multiply by -1

---------------

6j - k = 23

-6j - 3k = -11 (result of multiplying by -1)

---------------

-4k = 12

k = -12/4

k = - 3 <=====

6j - k = 23

6j - (-3) = 23

6j + 3 = 23

6j = 23 - 3

6j = 20

j = 20/6

j = 10/3 <===

solution is : j = 10/3 and k = -3

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Answer:

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Step-by-step explanation:

Let the store earned $x in December.

Therefore,

Money spent to buy new inventory =\frac{1}{4} x

Remaining money = x - \frac{1}{4} x =\frac{3}{4} x

Money used to pay bills =\frac{1}{2} \times \frac{3}{4} x=\frac{3}{8} x

Money still left over = $3,000

Total money earned in December = \frac{1}{4} x+ \frac{3}{8} x+3,000

\therefore x= \frac{1}{4} x+ \frac{3}{8} x+3,000

\therefore x= \frac{2}{8} x+ \frac{3}{8} x+3,000

\therefore x= \frac{5}{8} x+ 3,000

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2 years ago
If Logan has $30.67 and Jenny has $12.69 how much more does Logan have than Jenny
e-lub [12.9K]
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2 years ago
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3 years ago
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3 years ago
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