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Solution:
8x + 5y = 9 .....[1]
3x + 2y = 4....[2]
From equation [2], we obtain
3x + 2y = 4
3x = 4 - 2y
x = [4 - 2y]/3 ....[3]
Substituting x = [4 - 2y]/3 in equation [1], we obtain
8[[4 - 2y]/3] + 5y = 9
[32 - 16y + 15y]/3 = 9
32 - y = 27
y = 32 - 27
y = 5
Thus, x = [4 - 2 × 5]/3 [From equation[3]]
x = -6/3
x = - 2
Hence, x = - 2, y = 5
Again, by cross-multiplication method
8x + 5y = 9
3x + 2y = 4
8x + 5 - 9 = 0
3x + 2 - 4 = 0
a₁ = 8, b₁ = 5, c₁ = - 9
a₂ = 3, b₂ = 2, c₂ = - 4
[x/[b₁c₂ - b₂c₁] = y/[c₁a₂ - c₂a₁] = 1/[a₁b₂ - a₂b₁]]
x/[-20 - [-18]] = y/[-27 - [-32]] = 1/[16 - 15]
x/[-2] = y/5 = 1
x = - 2 and y = 5
☛ Check: NCERT Solutions Class 10 Maths Chapter 3
Video Solution:
Solve the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5y = 9; 3x + 2 y = 4
NCERT Solutions for Class 10 Maths - Chapter 3 Exercise 3.5 Question 3
Summary:
On solving the following pair of linear equations by the substitution and cross-multiplication methods: 8x + 5 y = 9 and 3x + 2 y = 4 we get x = - 2, and y = 5.
☛ Related Questions:
- Form the pair of linear equations in the following problems, and find their solutions [if they exist] by the elimination method: [i] If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction? [ii] Five years ago, Nuri was thrice as old as Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu? [iii] The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number. [iv] Meena went to a bank to withdraw ₹. 2000. She asked the cashier to give her ₹. 50 and ₹. 100 notes Meena got ₹. 25 notes in all. Find how many notes of ₹. 50 and ₹. 100 she received. [v] A lending library has a fixed charge for the first three days and an additional charge for each day Saritha paid ₹. 27 for a book kept for seven days, while Susy paid ₹. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
- Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using the cross multiplication method. [i] x - 3y - 3 = 0; 3x - 9 y - 2 = 0 [ii] 2x + y = 5; 3x + 2 y = 8 [iii] 3x - 5 y =-20; 6x - 10 y = 40 [iv] x - 3y - 7 = 0; 3x - 3y - 15 = 0
- [i] For which values of a and b will the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7 [a - b] x + [a + b] y = 3a + b - 2 [ii] For which value of k will the following pair of linear equations have no solution? 3x + y = 1 [2k -1] x + [k -1] y = 2k +1
Given pair of linear equations is
8x + 5y = 9 …[i]
And 3x + 2y = 4 …[ii]
On multiplying Eq. [i] by 2 and Eq. [ii] by 5 to make the coefficients of y equal, we get the equation as
16x + 10y = 18 …[iii]
15x + 10y = 20 …[iv]
On subtracting Eq. [iii] from Eq. [iv], we get
15x + 10y – 16x – 10y = 20 – 18
⇒ – x = 2
⇒ x = – 2
On putting x = – 2 in Eq. [ii], we get
3x + 2y = 4
⇒ 3[ – 2] + 2y = 4
⇒ – 6 + 2y = 4
⇒ 2y = 4 + 6
⇒ 2y = 10
y = 10/2 = 5
Hence, x = 2 and y = 5 , which is the required solution.
Given equation are 8x + 5y = 9 and 3x + 2y = 4
Comparing with a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0, We have
a1 = 8, b1 = 5, c1 = -9 and a2 = 3, b2 = 2, c2 = - 4
Now, x = `[ b_1c_2 - b_2c_1 ]/[ a_1b_2 - a_2b_1 ] and y = [ c_1a_2 - c_2a_1 ]/[ a_1b_2 - a_2b_1 ]`
⇒ x = `[ 5 xx [ - 4 ] - 2 xx [ - 9 ]]/[ 8 xx 2 - 3 xx 5 ] and y = [ - 9 xx 3 - [ - 4 ] xx 8 ]/[ 8 xx 2 - 3 xx 5 ]`
⇒ x = `[ - 20 + 18 ]/[ 16 - 15 ] and y = [ - 27 + 32 ]/[ 16 - 15 ]`
⇒ x = `-2/1 and y = 5/1`
⇒ x = - 2 and y = 5.