PLEASE ANSWER ALL QUESTIONS NEEP HELP HAVING VERY DIFFICULT

PLEASE ANSWER ALL QUESTIONS. NEEP HELP, HAVING VERY DIFFICULT TIME. THANK YOU!

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Solve the system of equations by the addition method. x+ 6y 5x-30 15 Select the correct choice below and if necessary, fill in the answer box to complete your choice O A. The solution is O B. ere are infinitely many solutions. O C. There is no solution.

Solution

1. We have –x+6y = -3…(1) and 5x -30y = 15 or, ondiving both the sides by 5, we get x-6y = 3…(2).

Now, on adding the 1^st and the 2^nd equations, we get –x+6y + x -6y = -3+3 or, 0 = 0. This situation arises because the two equations are the same. The 2^nd eqiation can be obtained by multiplying the 1^st equation by -1. The solution is 6y = -3 –x or y = -1/6(3+x). On substituting arbitrary infinite values of x , we will get infinite values of y. Thus, there are infinite solutions. Option B is the correct answer.

2. We have 6x+5y=-13…(1)and 3x-5y = 31…(2). On adding the two equations, we get 6x+5y+3x-5y = -13+31 or, 9x = 18 so that x = 18/9 =2. On substituting x = 2 in the 2^nd equation, we get 3*2-5y = 31 or, 5y = 6-31 = -25 so that y = -25/5 = -5. The answer is (x,y) = (2,-5). Option A.

3. We have 2x+4y = 0 or, on dividing both the sides by 2, we have x+2y = 0…(1) and 4x+8y = 8, or, on dividing both the sides by 4, we have x+2y = 2…(2). This means that 0 = x+2y = 2 which is apparently incorrect. Hence there is no solution as the given equations are inconsistent. Option C is the correct answer.

4. Let the larger number be x and the smaller number be y. Then x+y = 32…(1) and x-y =10…(2). On adding the two equations, we get x+y+x-y = 32+10 or 2x = 42 so that x= 42/2 = 21. On substituting x = 21 in the 2^nd equation, we get 21-y = 10 or, y = 21-10 = 11. Theus the larger number is 21 and the smaller number is 11.

5. We have y = 4x+1…(1) and 2y-4x = 14 or, 2y = 4x+14 or, on dividing both the sises by 2, we get y = 2x+7…(2). Now, on substituting y = 4x+1 from the 1^st equation into the 2^nd equation, we get 4x+1 = 2x+7 or, 4x-2x = 7-1 or, 2x = 6 so that x = 6/2 = 3. Then , on substituting x = 3 in the 1^st equation, we get y = 4*3+1 = 12+1 = 13. Thus, the solution is (x,y) = (3,13).

6. Let the1st and the 2^nd numbers be x and y respectively. Then

x = 5y-8…(1) and

x+y = 28…(2)

On substituting x = 5y-8 in the 2^nd equation, we get 5y-8+y = 28 or, 6y = 28+8 = 36 so that y = 36/6 = 6. Then x = 5*6-8 = 30-8 = 22. Thus, the smaller number is 6 and the larger number is 22.