Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
7.5k views
in Negative Numbers and Integers by (46.9k points)
closed by

Solve the following simultaneous equations using Cramer’s rule.

i. 6x – 4y = -12 ; 8x – 3y = -2

ii. 4m + 6n = 54 ; 3m + 2n = 28

iii. 2x + 3y = 2 ; x –\(\frac{y}{2}\)=\(\frac{1}{2}\)

1 Answer

+1 vote
by (47.6k points)
selected by
 
Best answer

i The given simultaneous equations are

6x – 4y = -12

∴ 3x – 2y = -6 …(i) [Dividing both sides by 2]

8x – 3y = -2 …(ii)

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2 , we get

a1 = 3, b1 = -2, c1 = -6 and

a2 = 8, b2 = -3, c2 = -2

∴ (x, y) = (2, 6) is the solution of the given simultaneous equations.

ii. The given simultaneous equations are

4m + 6n = 54

2m + 3n = 27 …(i) [Dividing both sides by 2]

3m + 2n = 28 …(ii)

Equations (i) and (ii) are in am + bn = c form. Comparing the given equations with
a1m + b1n = c and a2 m + b2 n = c2 , we get

a1 = 2, b1 = 3, c= 27 and

a2 = 3, b2 = 2, c2 = 28

∴ (m, n) = (6, 5) is the solution of the given simultaneous equations.

iii. The given simultaneous equations are

2x + 3y = 2 …(i)

x =\(\frac{y}{2}\) =\(\frac{1}{2}\)

∴ 2x – y = 1 …(ii) [Multiplying both sides by 2]

Equations (i) and (ii) are in ax + by = c form. Comparing the given equations with
a1x + b1y = c1 and a2x + b2y = c2 , we get

a1 = 2, b1 = 3, c1 = 2 and

a2 = 2, b2 = -1, c2 = 1

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...