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MA_775_DIABLO [31]
4 years ago
6

Combining like terms

Mathematics
1 answer:
Inessa05 [86]4 years ago
8 0

Combine like terms with common variable to simplify an expression.

1) (0b +10) + 5(b + 2) = 10 + 5b + 10 = 5b + 20

2) -7(n + 3) - 8(1 + 8n) = -7n -21 - 8 - 64n = -71n - 29


Use the distributive property to expand the expression.

1) -6(3x - 8) = -18x + 48

2) (-m + 9)(-3) = 3m - 27

3) 3(-6 - x) = -18 - 3x

4) 2(x - y + 1) = 2x - 2y + 2

5) 4(6p + 2q - 2p) = 24p + 8q - 8p = 16p + 8q

6) p(m + 2) = pm + 2p

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At a bargain store, Spongebob bought 3 items that each cost the same amount. Patrick bought 5 items that each cost the same amou
STALIN [3.7K]

Answer:

Cost of each item bought by Spongebob =x=\$3.75

Cost of items bought by Spongebob = 3x=3(3.75)=\$11.25

Cost of each item bought by Patrick =x-1.50=3.75-1.50=\$2.25

Cost of items bought by Patrick = 5(x-1.50)=5(3.75-1.50)=\$11.25

Step-by-step explanation:

Let x denotes cost of each item bought by Spongebob.

Number of items bought by Spongebob = 3

So,

Cost of items bought by Spongebob = 3x

Cost of each item bought by Patrick = x-1.50

Number of items bought by Patrick = 5

Cost of items bought by Patrick = 5(x-1.50)

Both Spongebob and Patrick paid the same amount of money.

So,

3x=5(x-1.50)\\3x=5x-7.5\\5x-3x=7.5\\2x=7.5\\x=\frac{7.5}{2} \\x=3.75

So,

Cost of each item bought by Spongebob =x=\$3.75

Cost of items bought by Spongebob = 3x=3(3.75)=\$11.25

Cost of each item bought by Patrick =x-1.50=3.75-1.50=\$2.25

Cost of items bought by Patrick = 5(x-1.50)=5(3.75-1.50)=\$11.25

6 0
3 years ago
This is due like right now, please help me!!!!
Varvara68 [4.7K]

\bold{\huge{\orange{\underline{ Solution }}}}

\bold{\underline{ Given \: Rules}}

  • <u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
  • <u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
  • <u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

\bold{\underline{ Let's \: Begin}}

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }

\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}

\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

\sf{ 158 + b = 166 + a, 156 + d = 166 + a }

\sf{ b = 166 - 158 + a,  d = 166 - 156 + a }

\sf{ b = 8 + a,  d = 10 + a ...(3)}

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

\sf{229+a = 161+8+a+c = 168+10+a+e}

\sf{ 229+a = 169+a+c = 178+a+e}

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}

\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}

\sf{ c = 60 \: and \: e = 51 }

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }

\sf{ 158 + b = 156 + d = 166 + a = 189}

<u>Now</u><u>, </u>

\sf{ For \: b,  158 + b = 189 }

\sf{ b = 189 - 158 }

\sf{ b = 31}

\sf{ For \: d ,  156 + b = 189 }

\sf{ d = 189 - 156 }

\sf{ d = 33}

\sf{ For \: a,  166 + a = 189 }

\sf{ a = 189 - 166 }

\sf{ a = 23 }

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

8 0
3 years ago
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