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Linear Equations

Linear Equations

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Linear Equations 1
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Linear Equations : Level 1
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CAT 2025 Lesson : Linear Equations - 2 Equations & 3 Variables

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4.7 2 equations for 3 variables

In these types we will be provided 222 equations with 333 variables. We will not be able to solve and find the values of each of the variables as we have one equation less than needed. Therefore, in these questions we will be asked to find the value of a dependent expression, which could possibly be deduced from the given 222 equations. This is shown in the examples below.

Example 15

If 222 nuts, 333 screws and 444 bolts cost Rs. 181818 and 777 nuts, 101010 screws and 131313 bolts cost Rs. 595959, then how much does a person need to pay to purchase 111 nut, 111 screw and 111 bolt?

Solution

Let n,sn, sn,s and bbb be the cost of 111 nut, 111 screw and 111 bolt respectively.

2n+3s+4b=182n + 3s + 4b = 182n+3s+4b=18 -----(111)
7n+10s+13b=597n + 10s + 13b = 597n+10s+13b=59 -----(222)

We note that if we subtract 333 times the first equation from the second, we get the total cost of 111 nut, 111 screw and 111 bolt.
Eq(222) −3×- 3 \times−3× Eq(111) ⇒ 7n+10s+13b−(6n+9s+12b)=59−54 7n + 10s + 13b - (6n + 9s + 12b) = 59 - 547n+10s+13b−(6n+9s+12b)=59−54
⇒ n+s+b=5 n + s + b = 5n+s+b=5

Answer: Rs. 555


In the last question it was easy to observe. Almost always in entrance tests, you should be able to solve these by observation. In questions where you are not sure if the expression you need to find is dependent on the other two, this approach will help.

Example 16

If 4x+3y+2z=404x + 3y + 2z = 404x+3y+2z=40 and 3x−y−2z=303x - y - 2z = 303x−y−2z=30, then
(I) 6x+11y+10z=6x + 11y + 10z =6x+11y+10z= ?
(II) 5x+11y+10z=5x + 11y + 10z =5x+11y+10z= ?

Solution

To find the dependent equation, we need to multiply the first equation by aaa and the second by bbb and find the values of aaa and bbb that form the coefficients of the variables in the dependent equation.

Case I

4x+3y+2z=404x + 3y + 2z = 404x+3y+2z=40 ----- ×a\times a×a
3x−y−2z=303x - y - 2z = 303x−y−2z=30 ----- ×b\times b×b
6x+11y+10z=6x + 11y + 10z =6x+11y+10z= ?

When the two equations are multiplies by constants aaa and bbb respectively, the third expression should form. So, each of the coefficients of x,yx, yx,y and zzz should match.

Eq(111): 4ax+3bx=6x4ax + 3bx = 6x4ax+3bx=6x ⇒ 4a+3b=64a + 3b = 64a+3b=6
Eq(222): 3ay−by=11y3ay - by = 11y3ay−by=11y ⇒ 3a−b=113a - b = 113a−b=11
Eq(333): 2az−2bz=10z2az - 2bz = 10z2az−2bz=10z ⇒ a−b=5a - b = 5a−b=5

Eq(222) - Eq(333) ⇒ 3a−a=11−5 3a - a = 11 - 53a−a=11−5 ⇒ a=3a = 3a=3
Substituting a=3a = 3a=3 in Eq(333), we get b=−2b = -2b=−2

Now, we need to check if a=3a = 3a=3 and b=−2b = -2b=−2 satisfies Eq (111).
4×3+3×−2=64 \times 3 + 3 \times -2 = 64×3+3×−2=6

4x+3y+2z=404x + 3y + 2z = 404x+3y+2z=40 ----- ×3\times 3×3
3x−y−2z=303x - y - 2z = 303x−y−2z=30 ----- ×−2\times -2×−2
⇒ 6x+11y+10z=60 \bm{6x + 11y + 10z = 60}6x+11y+10z=60

Case II:

If aaa and bbb are the multiples, then compared to Case I, Eq(111) is different. However, Eq(222) and Eq(333) are the same.
Eq(111): 4a+3b=54a + 3b = 54a+3b=5
Eq(242424): 3a−b=113a - b = 113a−b=11
Eq(333): a−b=5a - b = 5a−b=5

Like in Case I, solving Eq(222) and Eq(333), we get a = 333, b = −2-2−2.

With these values of aaa and bbb, we get 4a+3b=64a + 3b = 64a+3b=6. However, Eq(111) is 4a+3b=54a + 3b = 54a+3b=5

Therefore, 5x+11y+10z=?5x + 11y + 10z = ?5x+11y+10z=? is not dependent on the equations given in the question and, therefore, cannot be determined.

Answer:(I) 606060; (II) Cannot be determined


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