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Studentka2010 [4]
3 years ago
13

How do you do this problem?

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0
By the use of substitution:for x,

2x-3y=21 ;

will equal 2x-3 (3-x)=21

2x-9+3x=21

5x=30

x=6.

for y:

since y equals 3-x;

y=3-6=-3
astra-53 [7]3 years ago
5 0
By substitution is how to solve

2x-3y=21
y=3-x

2x-3(3-x)=21 
2x-9+3x=21
5x=30
x= 6

now substitute our x value into y=3-x
so we get 

y=3-6 = -3

so x=6 and y=-3  hope this helps!!!!! Thank you!
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VashaNatasha [74]

Answer:

a. Ameribank-$15,157.50

b. Capital Two-$4,646.25

Step-by-step explanation:

a. Tad's savings is $15,000, we calculate his total amount at the end of the year for each bank:

#Ameribank

A=P+I=P+PRT\\\\=15000+15000\times 0.0105\times 1\\\\=\$15,157.50

#Huffington( we use the effective rate to calculate the compound amount):

i_m=(1+i/m)^m-1\\\\=(1+0.0095/12)^[12}-1=0.009541\\\\A=P(1+i_m)^n\\\\=15000(1.009541)^1\\\\=\$15,143.12

#Sixth-Third, Take 1 yrs=52 weeks:

i_m=(1+i/m)^m-1\\\\=(1+0.01/52)^{52}-1=0.01005\\\\A=15000(1.01005)^1\\\\=\$15,150.74

#Hence, Ameribank is the best option as his money grows to $15,157.50 which is greater than all the remaining two options.

b. We use the compound interest formula A=P(1+r/n)^{nt} to determine which bank gives the best option:

#Capital Two. r=3.75%, n=12,t=4

A=P(1+r/n)^{nt}\\\\=4000(1+0.0375/12)^{12\times4}\\\\=\$4,646.25

#J.C Morgan, t=2, r=3.55% n=12

A=P(1+r/n)^{nt}\\\\=4000(1+0.0355/12)^{12\times 2}\\\\=\$4,293.87

#Silverman Slacks, n=12,t=3, r=3.65%

A=P(1+r/n)^{nt}\\\\=4000(1+0.0365/12)^{12\times3}\\\\=\$4,462.14

We compare the investment amounts after t years:

Capital>Silver>Morgan=4646.25>4462.14>4293.87

Hence, Capital two is the best option with an investment amount of $4,646.25

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Answer:

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Step-by-step explanation:

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