Question

10th class algebra
practice set1.1
qo no.2 solve the following simultaneous equation
1. 3a+5b=26; a +5b=22
2.x+7y=10; 3x-2y=7​

Answer 1

Step-by-step explanation:

1) Above is the answer

2) Above is the answer

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Answer 2

Answer:

a=2 and b=4

x=3 y=1

Step-by-step explanation:

1) 3a+5b=26------ eq 1

a +5b=22-----------eq 2

Subtracting eq 1 qnd 2, we get:

3a+5b-a-5b= 26-22

=> 2a= 4

=> a= 4/2= 2

substituting a= 2 in eq 2, we get

=> 2+5b= 22

=> 5b=22-2= 20

=> b= 20/5= 4

therefore a=2 and b=4

2)x+7y=10

x= 10-7y ----- eq 1

3x-2y=7​-------eq 2

substituting the value of x in eq 2 we get:

3(10-7y)- 2y=7

=> 30-21y-2y= 7

=> 30- 23y= 7

=> -23y= 7-30= -23

=> y=1

substituting the value of y in eq 1 we get;

x= 10- 7*1= 10-7=3

therefore x=3 and  y=1

hope this will help you