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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 22 equations with 33 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 22 equations. This is shown in the examples below.

Example 15

If 22 nuts, 33 screws and 44 bolts cost Rs. 1818 and 77 nuts, 1010 screws and 1313 bolts cost Rs. 5959, then how much does a person need to pay to purchase 11 nut, 11 screw and 11 bolt?

Solution

Let n,sn, s and bb be the cost of 11 nut, 11 screw and 11 bolt respectively.

2n+3s+4b=182n + 3s + 4b = 18 -----(11)
7n+10s+13b=597n + 10s + 13b = 59 -----(22)

We note that if we subtract 33 times the first equation from the second, we get the total cost of 11 nut, 11 screw and 11 bolt.
Eq(22) −3×- 3 \times Eq(11) ⇒ 7n+10s+13b−(6n+9s+12b)=59−54 7n + 10s + 13b - (6n + 9s + 12b) = 59 - 54
⇒ n+s+b=5 n + s + b = 5

Answer: Rs. 55


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

Solution

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

Case I

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

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

Eq(11): 4ax+3bx=6x4ax + 3bx = 6x ⇒ 4a+3b=64a + 3b = 6
Eq(22): 3ay−by=11y3ay - by = 11y ⇒ 3a−b=113a - b = 11
Eq(33): 2az−2bz=10z2az - 2bz = 10z ⇒ a−b=5a - b = 5

Eq(22) - Eq(33) ⇒ 3a−a=11−5 3a - a = 11 - 5 ⇒ a=3a = 3
Substituting a=3a = 3 in Eq(33), we get b=−2b = -2

Now, we need to check if a=3a = 3 and b=−2b = -2 satisfies Eq (11).
4×3+3×−2=64 \times 3 + 3 \times -2 = 6

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

Case II:

If aa and bb are the multiples, then compared to Case I, Eq(11) is different. However, Eq(22) and Eq(33) are the same.
Eq(11): 4a+3b=54a + 3b = 5
Eq(2424): 3a−b=113a - b = 11
Eq(33): a−b=5a - b = 5

Like in Case I, solving Eq(22) and Eq(33), we get a = 33, b = −2-2.

With these values of aa and bb, we get 4a+3b=64a + 3b = 6. However, Eq(11) is 4a+3b=54a + 3b = 5

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

Answer:(I) 6060; (II) Cannot be determined


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