Question
Solve the following pair of equations.
-2 x – 3 y = - 12 and 4 x - 3 y = 6
- 3, 4
- 2, 3
- 5, 6
- 3, 2
Hint:
The Linear Combination Method, aka The Addition Method , aka The Elimination Method. Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x-terms or the y -terms cancel out. Then solve for x (or y, whichever's left) and substitute back to get the other coordinate
The correct answer is: 3, 2
There are different methods for solving simultaneous linear Equations:
I. Elimination of a variable
II. Substitution
III. Cross-multiplication
IV. Evaluation of proportional value of variables
Here, we will be using Elimination of a variable method.
Here, we have 2 equations
-2x – 3y = - 12 can also be written as 2x + 3y = 12 (1)
4x - 3y = 6 (2)
Multiplying 2 in equation(1)
2 (2x + 3y = 12) = 4x +6y = 24 (3)
Subtracting equation 1 and 3 , we get :
4x + 6y = 24
-
4x –3y = 6
___________
= 9y = 18
y= 2
Substitute the value of y in equation (2), we get:
4x -3(2) = 6
x= 3
Thus, (x, y) = (3, 2)
Related Questions to study
Solve the following pair of equations.
![2 over x plus 3 over y equals 13 space a n d space 5 over x minus 4 over y equals negative 2](data:image/png;base64,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)
Solve the following pair of equations.
![2 over x plus 3 over y equals 13 space a n d space 5 over x minus 4 over y equals negative 2](data:image/png;base64,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)
Coach Jhon buys 20 bats and 5 balls for his team.
A ball costs x rupees and a bat costs y rupees. John spends a total of 400 rupees on these two items.
Express x in terms of y.
Coach Jhon buys 20 bats and 5 balls for his team.
A ball costs x rupees and a bat costs y rupees. John spends a total of 400 rupees on these two items.
Express x in terms of y.
2x +y = 10 Complete the missing value in the solution to the equation. (……..,- 6 )
2x +y = 10 Complete the missing value in the solution to the equation. (……..,- 6 )
James needs to solve the system of equations using elimination.
-3x + 5y = 15 and 2x – 5y = -15
What variable should James should solve first
James needs to solve the system of equations using elimination.
-3x + 5y = 15 and 2x – 5y = -15
What variable should James should solve first
Solve the system of equations using elimination.
3x + 2y = -13 and -3x+y= 25
Solve the system of equations using elimination.
3x + 2y = -13 and -3x+y= 25
3x – y =12
3x + 5y = -6
3x – y =12
3x + 5y = -6
-5x – 2y = 9
-2x + 3y = 15
-5x – 2y = 9
-2x + 3y = 15
4x – 5y = -5
4x – 4y = 0
4x – 5y = -5
4x – 4y = 0
4x – 4y = 4
x – 3y = 5
4x – 4y = 4
x – 3y = 5
x – 5y = 20
x – y = 4
x – 5y = 20
x – y = 4
-x – y = -5
-3x + 3y = -3
-x – y = -5
-3x + 3y = -3
3y = 15
-3x – 5y = -10
3y = 15
-3x – 5y = -10
-4x – 5y = -1
-2x – 5y = 7
-4x – 5y = -1
-2x – 5y = 7
-2x – 5y = 10
-x – y = -1
-2x – 5y = 10
-x – y = -1
-5x + y = -3
5x – 4y = 12
We can solve the equation in two variable using elimination method, substitution method and cross multiplication method. In elimination method we can find the solution of the two variables by cancelling one variable by adding, multiplying or subtracting both the equations from which we will get the value of one variable. After getting the value of one variable we can put the value of that variable in equation and get the value of another variable.
-5x + y = -3
5x – 4y = 12
We can solve the equation in two variable using elimination method, substitution method and cross multiplication method. In elimination method we can find the solution of the two variables by cancelling one variable by adding, multiplying or subtracting both the equations from which we will get the value of one variable. After getting the value of one variable we can put the value of that variable in equation and get the value of another variable.