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Goshia [24]
3 years ago
9

Put the equation in standard form with a positive x coefficient. 2y+5=-4(x+2)

Mathematics
1 answer:
umka21 [38]3 years ago
6 0

The standard form of equation is: 4x+2y = -13

Step-by-step explanation:

the standard form of a linear equation in two variables is given by:

ax+by = c

Given equation is:

2y+5=-4(x+2)

Distributing -4 into the round bracket

2y+5 = -4x-8

Adding 4x on both sides

4x+2y+5 = -4x-8+4x\\4x+2y+5 = -8

subtracting 5 from both sides

4x+2y+5-5 = -8-5\\4x+2y = -13

The standard form of equation is: 4x+2y = -13

Keywords: Linear equation, variables

Learn more about linear equations at:

  • brainly.com/question/6112033
  • brainly.com/question/6108628

#LearnwithBrainly

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There is a single sequence of integers $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, $a_7$ such that \[\frac{5}{7} = \frac{a_2}{2!} + \frac
Nataliya [291]

You have a single sequence of integers a_2,\ a_3,\ a_4,\ a_5,\ a_6,\ a_7 such that

\dfrac{a_2}{2!} + \dfrac{a_3}{3!} + \dfrac{a_4}{4!} + \dfrac{a_5}{5!} + \dfrac{a_6}{6!} + \dfrac{a_7}{7!}=\dfrac{5}{7},

where 0 \le a_i < i for i = 2, 3, \dots, 7.

1. Multiply by 7! to get

\dfrac{7!a_2}{2!} + \dfrac{7!a_3}{3!} + \dfrac{7!a_4}{4!} + \dfrac{7!a_5}{5!} + \dfrac{7!a_6}{6!} + \dfrac{7!a_7}{7!}=\dfrac{7!\cdot 5}{7},\\ \\7\cdot 6\cdot 5\cdot 4\cdot 3\cdoa a_2+7\cdot 6\cdot 5\cdot 4\cdot a_3+7\cdot 6\cdot 5\cdot a_4+7\cdot 6\cdot a_5+7\cdot a_6+a_7=6!\cdot 5,\\ \\7(6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)+a_7=3600.

By Wilson's theorem,

a_7+7\cdot (6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6)\equiv 2(\mod 7)\Rightarrow a_7=2.

2. Then write a_7 to the left and divide through by 7 to obtain

6\cdot 5\cdot 4\cdot 3\cdoa a_2+6\cdot 5\cdot 4\cdot a_3+6\cdot 5\cdot a_4+6\cdot a_5+a_6=\dfrac{3600-2}{7}=514.

Repeat this procedure by \mod 6:

a_6+6(5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5)\equiv 4(\mod 6)\Rightarrow a_6=4.

And so on:

5\cdot 4\cdot 3\cdoa a_2+ 5\cdot 4\cdot a_3+5\cdot a_4+a_5=\dfrac{514-4}{6}=85,\\ \\a_5+5(4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4)\equiv 0(\mod 5)\Rightarrow a_5=0,\\ \\4\cdot 3\cdoa a_2+ 4\cdot a_3+a_4=\dfrac{85-0}{5}=17,\\ \\a_4+4(3\cdoa a_2+ a_3)\equiv 1(\mod 4)\Rightarrow a_4=1,\\ \\3\cdoa a_2+ a_3=\dfrac{17-1}{4}=4,\\ \\a_3+3\cdot a_2\equiv 1(\mod 3)\Rightarrow a_3=1,\\ \\a_2=\dfrac{4-1}{3}=1.

Answer: a_2=1,\ a_3=1,\ a_4=1,\ a_5=0,\ a_6=4,\ a_7=2.

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3 years ago
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