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Solve the following pair of equations by cross multiplication method. 5x − 2y + 9 = 0, 4x + 3y = 2 - Mathematics

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प्रश्न

Solve the following pair of equations by cross multiplication method.

5x − 2y + 9 = 0, 4x + 3y = 2

योग
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उत्तर

Given equations:

5x − 2y + 9 = 0     ...(1)

4x + 3y = 2

4x + 3y − 2 = 0     ...(2)

Let’s write equations in standard form:

a1x + b1y1 + c1 = 0

a2x + b2y + c2 = 0

Here, they are in the form of,

a1 = 4, b1 = −2, c1 = 9

a2 = 4, b2 = 3, c2 = −2

Using the identity:

`x/(b_1c_2 - b_2c_1) = y/(c_1a_2 - c_2a_1) = 1/(a_1b_2 - a_2b_1)`

Now, substituting the values,

⇒ b1c2 − b2c1

= (−2)(−2) − (3)(9)

= 4 − 27

∴ b1c2 − b2c1 = −23

⇒ c1a2 − c2a1

= (9)(4) − (−2)(5)

= 36 + 10

∴ c1a2 − c2a1 = 46

⇒ a1b2 − a2b1

= (5)(3) − (4)(−2)

= 15 + 8

∴ a1b2 − a2b1 = 23

So, the value becomes,

`x/-23 = y/46 = 1/23`

Hence, finding x and y,

`x/-23 = 1/23`

`x = (-23)(1/23)`

`x = (-23)/23`

∴ x = −1

`y/46 = 1/23`

`y = (46)(1/23)`

`y = 46/23`

∴ y = 2

Thus, solving equations 5x − 2y + 9 = 0, 4x + 3y = 2 by cross multiplication method we get, x = −1 and y = 2.

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अध्याय 5: Simultaneous Linear Equations - EXERCISE 5A [पृष्ठ ५३]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
अध्याय 5 Simultaneous Linear Equations
EXERCISE 5A | Q II. (ii) | पृष्ठ ५३
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