what to do next. find value of x and y
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3x + y = 40 --------(1)
4x + 2y = 50 -------(2)
On multiplying equation (1) by 2, we get
6x + 2y = 80 -------(3)
On subtracting equation 2 from 3, we get
2x = 30
=> x = 15
Now,
On substituting the value of x in equation 1, we get
3 (15) + y = 40
=> 45 + y = 40
=> y = - 5
4x + 2y = 50 -------(2)
On multiplying equation (1) by 2, we get
6x + 2y = 80 -------(3)
On subtracting equation 2 from 3, we get
2x = 30
=> x = 15
Now,
On substituting the value of x in equation 1, we get
3 (15) + y = 40
=> 45 + y = 40
=> y = - 5
Answered by
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HEY THERE ..!!
HERE IS YOURS SOLUTION;
◆ 3x + y = 40
◆ 4x + 2y = 50
LETS BEGIN;
◆ 3x + y - 40 = 0 ------1)
◆ 4x + 2y - 50 = 0 --------2)
Here we us Elimination Method to get values of x & y.
2(3x + y - 40 = 0)
1(4x + 2y - 50 = 0)
【 Multipled by the coefficient of y】
=> 6x + 2y - 80 = 0
=> 4x + 2y - 50 = 0
Now, we subtract;
2x - 30 = 0
=> x = 30 ÷ 2
=> x = 15
Put x in eq" 1)
=> 3x + y - 40 = 0 ------1)
=> 3 (15) + y - 40 = 0
=> 45 + y - 40 = 0
=> 5 + y = 0
=> y = -5
★So, the value of x = 15 & y = -5★
↓IF ANY DOUBTS JUST COMMENT DOWN↓
KEEP CALM & BE BRAINLY
HOPE IT HELPS
HERE IS YOURS SOLUTION;
◆ 3x + y = 40
◆ 4x + 2y = 50
LETS BEGIN;
◆ 3x + y - 40 = 0 ------1)
◆ 4x + 2y - 50 = 0 --------2)
Here we us Elimination Method to get values of x & y.
2(3x + y - 40 = 0)
1(4x + 2y - 50 = 0)
【 Multipled by the coefficient of y】
=> 6x + 2y - 80 = 0
=> 4x + 2y - 50 = 0
Now, we subtract;
2x - 30 = 0
=> x = 30 ÷ 2
=> x = 15
Put x in eq" 1)
=> 3x + y - 40 = 0 ------1)
=> 3 (15) + y - 40 = 0
=> 45 + y - 40 = 0
=> 5 + y = 0
=> y = -5
★So, the value of x = 15 & y = -5★
↓IF ANY DOUBTS JUST COMMENT DOWN↓
KEEP CALM & BE BRAINLY
HOPE IT HELPS
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