Advertisements
Advertisements
प्रश्न
Solve the following pair of linear (simultaneous) equation by the method of elimination by substitution:
0.2x + 0.1y = 25
2(x - 2) - 1.6y = 116
Advertisements
उत्तर
The given pair of linear equations are
0.2x + 0.1y = 25 ...(1)
2(x - 2) - 1.6y = 116 ...(2)
Consider equation (1)
0.2x + 0.1y = 25
⇒ `(0.2x)/10+(0.1y)/10=25`
⇒ `(2x+y)/10=25`
⇒ 2x + y = 250
⇒ y = 250 - 2x ...(3)
Putting the value of y in equation (2)
⇒ 2(x - 2) - 1.6(250 - 2x) = 116
⇒ 2x - 4 - 400 + 3.2x = 116
⇒ 5.2x - 404 = 116
⇒ 5.2x = 116 + 404
⇒ 5.2x = 520
⇒ x = `520/5.2`
⇒ x = 100
From equation in (3)
`0.2/10(100) + 0.1y = 25`
`20+y/10=25`
`y/10=25-20`
`y/10=5`
y = 50
APPEARS IN
संबंधित प्रश्न
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) equation by the method of elimination by substitution:
2x + 3y = 8
2x = 2 + 3y
Solve the following pair of linear (simultaneous) equation using method of elimination by substitution:
3x + 2y =11
2x - 3y + 10 = 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
Solve the following simultaneous equations by the substitution method:
5x + 4y - 23 = 0
x + 9 = 6y
Solve the following simultaneous equations by the substitution method:
7x - 3y = 31
9x - 5y = 41
Solve the following simultaneous equations by the substitution method:
3 - (x + 5) = y + 2
2(x + y) = 10 + 2y
Solve the following simultaneous equations by the substitution method:
7(y + 3) - 2(x + 2) = 14
4(y - 2) + 3(x - 3) = 2
Solve the following pairs of equations:
`(6)/(x + y) = (7)/(x - y) + 3`
`(1)/(2(x + y)) = (1)/(3( x - y)`
Where x + y ≠ 0 and x - y ≠ 0
If a number is thrice the other and their sum is 68, find the numbers.
The age of the father is seven times the age of the son. Ten years later, the age of the father will be thrice the age of the son. Find their present ages.
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/10 + y/5` = 14, `x/8 + y/6` = 15
Solve by the method of elimination
`4/x + 5y` = 7, `3/x + 4y` = 5
Solve by the method of elimination
13x + 11y = 70, 11x + 13y = 74
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
