q20-23 with steps pls

Accepted Solution

20: Let [tex]x[/tex] be the amount of money. If this amount is shared between 8 people, each person will get x/8 dollars. But there actually are 6 people, so everyone is getting x/6 dollars. We know that this difference results in 30 dollars more, so we have [tex]\dfrac{x}{6}=\dfrac{x}{8}+30 \iff \dfrac{x}{6}-\dfrac{x}{8}=30[/tex] Rearrange the left hand side as [tex]\dfrac{4x-3x}{24}=30[/tex] And multiply both sides by 24 to get [tex]x=720[/tex] 21: Let [tex]M[/tex] and [tex]J[/tex] be the number of marbles owned by Mike and Judy, respectively. At the beginning, we have [tex]M=3J+11[/tex] If they both get 9 more marbles, Mike will have [tex]M+9[/tex] marbles, and Judy will have [tex]J+9[/tex] marbles. So, we have [tex]M+J+18=93 \iff M+J=75 \iff M=75-J[/tex] Plug this value in the first equation and we have [tex]75-J=3J+11 \iff 4J=64 \iff J=16[/tex] And we deduce [tex]M=16\cdot 3 + 11=59[/tex] So, the difference is  [tex]M-J=59-16=43[/tex] 22: Let [tex]x,y,z[/tex] be the number of $10, $20 and $50 coupon, respectively. We're given: [tex]\begin{cases}x=2y+3\\z=\frac{1}{2}y\\x+y+z=38\end{cases} We can write the third equation by substituting the expressions for x and z: [tex]x+y+z=(2y+3)+y+\left(\dfrac{1}{2}y\right)=38[/tex] Rearrange as follows: [tex](2y+3)+y+\left(\dfrac{1}{2}y\right)=38 \iff \dfrac{7}{2}y=35 \iff 7y=70 \iff y=10[/tex] And now we can deduce the number of the other coupons: [tex]x=2\cdot 10+3=23,\quad z=\dfrac{1}{2}\cdot 10=5[/tex] So, the total value is [tex]23\cdot 10+10\cdot 20+5\cdot 50=230+200+100=530[/tex] 23:  a) In order to find the first three terms of the sequence, you just need to plug n=1,2,3:[tex]a_1=4\cdot 1+3=7,\quad a_2=4\cdot 2+3=11,\quad a_3=4\cdot 3+3=15[/tex]b) We have[tex]a_r = 4r+3=71 \iff 4r=68 \iff r=\dfrac{68}{4}=17[/tex]c) 105 is a term of the sequence if and only if there exists an integer k such that[tex]a_k=4k+3=105 \iff 4k = 102 \iff k=\dfrac{102}{4}=25.5[/tex]So, 105 is not a term of the sequence.