Advertisements
Advertisements
Question
Solve by the method of elimination
`x/10 + y/5` = 14, `x/8 + y/6` = 15
Advertisements
Solution
`x/10 + y/5` = 14
L.C.M of 10 and 5 is 10
Multiply by 10
x + 2y = 140 → (1)
`x/8 + y/6` = 15
L.C.M of 8 and 6 is 24
3x + 4y = 360 → (2)
(1) × 2 ⇒ 2x + 4y = 280 → (3)
(2) × 1 ⇒ 3x + 4y = 360 → (2)
(3) – (2) ⇒ – x + 0 = – 80
∴ x = 80
Substitute the value of x = 80 in (1)
x + 2y = 140
80 + 2y = 140
2y = 140 – 80
2y = 60
y = `60/2`
y = 30
∴ The value of x = 80 and y = 30
APPEARS IN
RELATED QUESTIONS
Solve the pair of linear (simultaneous) equations by the method of elimination by substitution:
8x + 5y = 9
3x + 2y = 4
Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
2x - 3y = 7
5x + y= 9
Solve the pair of linear (simultaneous) equations by the method of elimination by substitution:
2x + 3y = 8,
2x = 2 + 3y
Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
1.5x + 0.1y = 6.2
3x - 0.4y = 11.2
Solve the following pair of linear (Simultaneous ) equation using method of elimination by substitution :
2( x - 3 ) + 3( y - 5 ) = 0
5( x - 1 ) + 4( y - 4 ) = 0
Solve the following pair of linear (simultaneous) equation using method of elimination by substitution :
2x - 3y + 6 = 0
2x + 3y - 18 = 0
Solve the following simultaneous equations by the substitution method:
2x + y = 8
3y = 3 + 4x
The ratio of passed and failed students in an examination was 3 : 1. Had 30 less appeared and 10 less failed, the ratio of passes to failures would have been 13 : 4. Find the number of students who appeared for the examination.
Solve by the method of elimination
x – y = 5, 3x + 2y = 25
Five years ago, a man was seven times as old as his son, while five year hence, the man will be four times as old as his son. Find their present age
