Advertisements
Advertisements
Question
Solve the following systems of equations:
`x + y/2 = 4`
`x/3 + 2y = 5`
Advertisements
Solution
The given system of equation is
`x + y/2 = 4` .......(i)
`x/3 + 2y = 5` ....(ii)
From (i), we get
`(2x + y)/2 = 4`
2x + y = 8
y = 8 - 2x
From (ii), we get
x + 6y = 15 ....(iii)
Substituting y = 8 - 2x in (iii), we get
x + 6(8 - 2x) = 15
`=> x + 48 - 12x = 15`
=> -11x = 15 - 48
=> -11x = -33
`=> x = (-33)/(-11) = 3`
Putting x = 3 in y = 8 - 2x we get
y = 8 - 2 x 3
= 8 - 6
= 2
y = 2
Hence, solution of the given system of equation is x= 3, y = 2
RELATED QUESTIONS
A train travels at a certain average speed for a distance of 54 km and then travels a distance of 63 km at an average speed of 6 km/h more than the first speed. If it takes 3 hours to complete the total journey, what is its first speed?
Solve the following system of equations by the method of cross-multiplication:
`\frac{x}{a} + \frac{y}{b} = a + b; \frac{x}{a^{2}} + \frac{y}{b^{2}} = 2`
Solve the following system of equations in x and y by cross-multiplication method
`(a – b) x + (a + b) y = a^2 – 2ab – b^2`
`(a + b) (x + y) = a^2 + b^2`
Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :
A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solve the following system of equations:
`x/3 + y/4 =11`
`(5x)/6 - y/3 = -7`
Solve the following system of equations:
0.5x + 0.7y = 0.74
0.3x + 0.5y = 0.5
Solve the following system of equations by the method of cross-multiplication:
`(x + y)/(xy) = 2`
`(x - y)/(xy) = 6`
Solve the following system of equations by the method of cross-multiplication:
`ax + by = (a + b)/2`
3x + 5y = 4
Solve each of the following systems of equations by the method of cross-multiplication :
2ax + 3by = a + 2b
3ax + 2by = 2a + b
Solve each of the following systems of equations by the method of cross-multiplication :
`x(a - b + (ab)/(a - b)) = y(a + b - (ab)/(a + b))`
`x + y = 2a^2`
Solve the following system of equations by the method of cross-multiplication:
mx – ny = m2 + n2
x + y = 2m
Solve each of the following systems of equations by the method of cross-multiplication :
`b/a x + a/b y - (a^2 + b^2) = 0`
x + y - 2ab = 0
Solve the system of equations by using the method of cross multiplication:
x + 2y + 1 = 0, 2x – 3y – 12 = 0
Solve the system of equations by using the method of cross multiplication:
`x/6 + y/15 - 4 = 0, x/3 - y/12 - 19/4 = 0`
Solve the system of equations by using the method of cross multiplication:
`(ax)/b - (by)/a - (a + b) = 0, ax - by - 2ab = 0`
Complete the activity to find the value of x.
3x + 2y = 11 ...(i) and 2x + 3y = 4 ...(ii)
Solution:
Multiply equation (i) by ______ and equation (ii) by ______.
`square` × (3x + 2y = 11) ∴ 9x + 6y = 33 ...(iii)
`square` × (2x + 3y = 4) ∴ 4x + 6y = 8 ...(iv)
Subtract (iv) from (iii),
`square` x = 25
∴ x = `square`
Solve the following pair of equations:
x + y = 3.3, `0.6/(3x - 2y) = -1, 3x - 2y ≠ 0`
Solve the following pair of equations:
`4x + 6/y = 15, 6x - 8/y = 14, y ≠ 0`
