Скачать или смотреть Q.3 | Exercise 3.5 | X Mathematics | NCERT | CBSE | solve by substitution and cross multiplication

Q.3 | Exercise 3.5 | X Mathematics | NCERT | CBSE | solve by substitution and cross multiplication methods

Q.3 Solve the following pair of linear equations by the substitution and cross-multiplication methods :

8𝒙+𝟓𝒚=𝟗 …(1)
𝟑𝒙+𝟐𝒚=𝟒 …(2)
Solution:
By Substitution Method:
From eq.(2), we have
𝒚=(𝟒−𝟑𝒙)/𝟐
Substituting this value in eq.(1),
8𝒙+𝟓×((𝟒−𝟑𝒙)/𝟐)=𝟗
(𝟏𝟔𝒙+𝟓(𝟒−𝟑𝒙))/𝟐=𝟗
(𝟏𝟔𝒙+𝟐𝟎−𝟏𝟓𝒙)/𝟐=𝟗
𝒙+𝟐𝟎=𝟏𝟖
𝒙=−𝟐
Substituting this value in eq.(2)
𝟑×(−𝟐)+𝟐𝒚=𝟒
𝟐𝒚=𝟒+𝟔
𝒚=𝟓
∴𝒙=−𝟐 𝒂𝒏𝒅 𝒚=𝟓

Now, By Cross-Multiplication Method:

Comparing the given equations by the standard form, we get
𝑎_1=8,𝑏_1=5 and 𝑐_1=−9
𝑎_2=3,𝑏_2=2 and 𝑐_2=−4

𝑥/(𝑏_1 𝑐_2−𝑏_2 𝑐_1 )=𝑦/(𝑐_1 𝑎_2−𝑐_2 𝑎_1 )=1/(𝑎_1 𝑏_2−𝑎_2 𝑏_1 )
𝑥/(5×(−4)−2×(−9))=𝑦/((−9)×3−(−4)×8)=1/(8×2−3×5)
𝑥/(−20+18)=𝑦/(−27+32)=1/(16−15)
𝑥/(−2)=𝑦/5=1/1

Now,
𝑥/(−2)=1/1
𝑥=−2
And,
𝑦/5=1/1
𝑦=5

∴𝒙=−𝟐 𝒂𝒏𝒅 𝒚=𝟓



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