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Mia has a certain amount of money. If she buys 4 pens and 1 pencil, she will have \$5 left over. If she buys 2 pens and 2 pencils, she will have \$3 left over. If Edward arrives with the same amount of money as Mia, together they can buy 7 pens and 4 pencils and spend all their money. If \(a\) is the cost of one pen and \(b \) is the cost of one pencil, compute the ordered pair \((a,b) \).

Sep 9, 2018
edited by Guest  Sep 9, 2018

#1
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Note: This is a long and detailed answer!

Let the amount of money that Mia has =M=same amount of money as Edward has
M - [4a + b] =5
M - [2a + 2b]=3
[7a + 4b] =2M

Solve the following system:
{-4 a - b + M = 5 | (equation 1)
-2 a - 2 b + M = 3 | (equation 2)
7 a + 4 b = 2 M | (equation 3)

Express the system in standard form:
{-(4 a) - b + M = 5 | (equation 1)
-(2 a) - 2 b + M = 3 | (equation 2)
7 a + 4 b - 2 M = 0 | (equation 3)

Swap equation 1 with equation 3:
{7 a + 4 b - 2 M = 0 | (equation 1)
-(2 a) - 2 b + M = 3 | (equation 2)
-(4 a) - b + M = 5 | (equation 3)

Add 2/7 × (equation 1) to equation 2:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a - (6 b)/7 + (3 M)/7 = 3 | (equation 2)
-(4 a) - b + M = 5 | (equation 3)

Multiply equation 2 by 7/3:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a - 2 b + M = 7 | (equation 2)
-(4 a) - b + M = 5 | (equation 3)

Add 4/7 × (equation 1) to equation 3:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a - 2 b + M = 7 | (equation 2)
0 a+(9 b)/7 - M/7 = 5 | (equation 3)

Multiply equation 3 by 7:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a - 2 b + M = 7 | (equation 2)
0 a+9 b - M = 35 | (equation 3)

Swap equation 2 with equation 3:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a+9 b - M = 35 | (equation 2)
0 a - 2 b + M = 7 | (equation 3)

Add 2/9 × (equation 2) to equation 3:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a+9 b - M = 35 | (equation 2)

0 a+0 b+(7 M)/9 = 133/9 | (equation 3)

Multiply equation 3 by 9/7:

{7 a + 4 b - 2 M = 0 | (equation 1)
0 a+9 b - M = 35 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

Add equation 3 to equation 2:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a+9 b+0 M = 54 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

Divide equation 2 by 9:
{7 a + 4 b - 2 M = 0 | (equation 1)
0 a+b+0 M = 6 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

Subtract 4 × (equation 2) from equation 1:
{7 a + 0 b - 2 M = -24 | (equation 1)
0 a+b+0 M = 6 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

Add 2 × (equation 3) to equation 1:
{7 a+0 b+0 M = 14 | (equation 1)
0 a+b+0 M = 6 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

Divide equation 1 by 7:
{a+0 b+0 M = 2 | (equation 1)
0 a+b+0 M = 6 | (equation 2)
0 a+0 b+M = 19 | (equation 3)

a = \$2,                b=\$6,                M=\$19

Sep 9, 2018
#2
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M - [4a + b] =5.............(1)
M - [2a + 2b]=3...........(2)
[7a + 4b] =2M.............(3)

M=5 +  4a + b -rearrange(1).....................(4)
M=3 + 2a + 2b -rearrange(2) and subt.....(5)
0 =2 + 2a - b.............................................(6)
2M=0 + 7a + 4b - rearrange (3)
2M=10 + 8a + 2b - Multiply (4) by 2 and subt.
0 =-10 - a + 2b.........................................(7)
Solve (6) and (7) as 2 simult.equations.
a=\$2   and   b=\$6. Sub these into (1)
M =\$19

Sep 10, 2018