Advertisements
Advertisements
Question
Solve by the method of elimination
`4/x + 5y` = 7, `3/x + 4y` = 5
Advertisements
Solution
Let `1/x` = a
4a + 5y = 7 → (1)
3a + 4y = 5 → (2)
(1) × 4 ⇒ 16a + 20y = 28 →(3)
(2) × 5 ⇒ 15a + 20y = 25 → (4)
(3) – (4) ⇒ a + 0 = 3
a = 3
Substitute the value of a = 3 in (1)
4(3) + 5y = 7
5y = 7 – 12
5y = −5
5y = `(-5)/5` = −1
But `1/x` = a
`1/x` = 3
3x = 1 ⇒ x = `1/3`
The value of x = `1/3` and y = −1
APPEARS IN
RELATED QUESTIONS
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 pairs of linear (simultaneous) equation using method of elimination by substitution:
`x/6 + y/15 = 4`
`x/3 - y/12 = 4 3/4`
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.
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.
In a ABC, ∠A = x°, ∠B = (2x - 30)°, ∠C = y° and also, ∠A + ∠B = one right angle. Find the angles. Also, state the type of this triangle.
Samidha and Shreya have pocket money Rs.x and Rs.y respectively at the beginning of a week. They both spend money throughout the week. At the end of the week, Samidha spends Rs.500 and is left with as much money as Shreya had in the beginning of the week. Shreya spends Rs.500 and is left with `(3)/(5)` of what Samidha had in the beginning of the week. Find their pocket money.
Solve by the method of elimination
2x – y = 3, 3x + y = 7
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
