Advertisements
Advertisements
Question
Solve the following pairs of linear (simultaneous) equation using method of elimination by substitution:
`x/6 + y/15 = 4`
`x/3 - y/12 = 4 3/4`
Advertisements
Solution
`x/6 + y/15 = 4`
⇒ `[ 5x + 2y ]/30 = 4`
⇒ 5x + 2y = 120
⇒ 5x = 120 - 2y
⇒ x = `[ 120 - 2y ]/5` ....(1)
And,
`x/3 - y/12 = 4 3/4`
⇒ `1/3( x - y/4 ) = 19/4`
⇒ `1/3([120 - 2y]/5 - y/4 ) = 19/4`
⇒ `[ 480 - 8y - 5y ]/20 = 57/4`
⇒ `[ 480 - 13y ]/20 = 57/4`
⇒ 480 - 13y = 285
⇒ 13y = 195
⇒ y = 15
Substituting the value of y in (1), we have
`x = [ 120 - 2 xx 15]/5 = [120 - 30]/5 = 90/5 = 18`
∴ Solution is x = 18 and y = 15
APPEARS IN
RELATED QUESTIONS
Solve the pair of linear (simultaneous) equation by the method of elimination by substitution :
2x - 3y = 7
5x + y= 9
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 :
2x - 3y + 6 = 0
2x + 3y - 18 = 0
Solve th following pair of linear (Simultaneous ) equation using method of elimination by substitution :
`[ 2x + 1]/7 + [5y - 3]/3 = 12`
`[3x + 2 ]/2 - [4y + 3]/9 = 13`
Solve the following simultaneous equations by the substitution method:
7x - 3y = 31
9x - 5y = 41
Solve the following simultaneous equations by the substitution method:
13 + 2y = 9x
3y = 7x
Solve the following simultaneous equations by the substitution method:
0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Solve the following simultaneous equations by the substitution method:
7(y + 3) - 2(x + 2) = 14
4(y - 2) + 3(x - 3) = 2
The difference of two numbers is 3, and the sum of three times the larger one and twice the smaller one is 19. Find the two numbers.
The sum of four times the first number and three times the second number is 15. The difference of three times the first number and twice the second number is 7. Find the numbers.
A two-digit number is such that the ten's digit exceeds thrice the unit's digit by 3 and the number obtained by interchanging the digits is 2 more than twice the sum of the digits. Find the number.
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
3(2x + y) = 7xy, 3(x + 3y) = 11xy
Solve by the method of elimination
`4/x + 5y` = 7, `3/x + 4y` = 5
