Forming a pair of simultaneous equations-find two numbers with a sum of 15 and a difference of 4?

twice one number added to three times another gives 21. find the numbers, if the difference between them is 3.Forming a pair of simultaneous equations-find two numbers with a sum of 15 and a difference of 4?
For both questions, what you need to do is call one number x and the other y. Then you write out equations in terms of x and y relating them. The final step in each case is then to solve for x and y.





1)





x + y = 15 ...A


x - y = 4 ...B





2x = 19 ...Get this by adding equation A to B





x = 19/2





y = x - 4 = 11/2





2)





2x + 3y = 21 ...A


x - y = 3 ...B





Multiply equation B by three:





3x - 3y = 9 ...C





Add equations C and A





5x = 30 =%26gt; x = 6





From B: y = x - 3 =%26gt; y = 3Forming a pair of simultaneous equations-find two numbers with a sum of 15 and a difference of 4?
1-a # 2-b


1st equation- a+b=15


2nd equation- a-b=4


1st + 2nd 2a= 19


a= 9.5


9.5+ b =15 (15-9.5 = 5.5)


b= 5.5


1st- 9.5+ 5.5= 15


2nd- 9.5-5.5=4


Got it? the numbers are 9.5 and 5.5





1st equation- 2a+3b=21


2nd equation- a-b=3 *2


3rd equation- 2a-2b=6


1st-3rd- 5b=15 /5


b= 3


2a+9=21 (21-9 = 12) /2


a= 6


1st- 12+18=21


2nd 6-3= 3


Right? a=6 b=3


thanks I have a test on that this friday :) good practice
Smaller no. (y):


2x + 3y = 21


y = (21 - 2x)/3





x - y = 3


y = x - 3





Larger no. (x):


21 - 2x = 3(x - 3)


21 - 2x = 3x - 9


5x = 30


x = 6





Smaller no. (1st equation):


= (21 - 2[6])/3


= (21 - 12)/3


= 9/3 or 3





Answer: 6 %26amp; 3 are the numbers.





Proof (twice one no. added to 3 times another gives 21):


= 2(6) + 3(3)


= 12 + 9


= 21





Proof (difference between them is 3):


= 6 - 3


= 3
9.5 and 5.5





2x+3y=21


lx-yl=3





x=6


y=3