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
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Step-by-step explanation:
1) Above is the answer
2) Above is the answer
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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
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