基本方程式-解決例

Q 1-8x + 5y = 9および3x + 2y = 4の場合、yとは何ですか？

A -5

B -6

C -7

D -8

Answer - A

Explanation

The given equations are:
8x+5y = 9 ...(a)
3x+2y = 4 ...(b)
On multiplying (a) by 2, (b) by 5 and subtracting, we get: x= -2
Putting x = -2 in (b), we get:
-6 + 2y = 4 => 2y = 10 ∴ y = 5

Q 2-5 / x + 6y = 13および3 / x + 4 = 7の場合、yの値を見つけます。

A -0

B --1

C --- 2

D --3

Answer - C

Explanation

The given equation is:
5/x +6y = 13 ...(a)
3/x+4y =7 ...(b)
On multiplying (a) by 3, (b) by 5 and subtracting, we get:
-2y = 4 ∴ y = -2

Q 3-与えられた（x + y-8）/ 2 =（x + 2y-14）/ 3 =（3x + y-12）/ 11。次にx、yは

A -1,7

B -2,7

C -2,6

D -1,5

Answer - D

Explanation

Taking first two parts, we get:  
(x+y-8)/2 = (x+2y-14)/3   
=> 3 (x+y-8) = 2(x+ 2y-14)
=> 3x+3y-24 =   2x+4y -28
=> x- y= -4 ...(1)
Taking last two parts, we get:
(x+2y-14)/3 = (3x+y-12)/11 
=> 11 (x+2y-14) = 3(3x+y-12)
=> 11x+ 22y - 154 = 9x+3y -36
=> 2x+19y- 118 ...(2)
Multiplying (1) by 2 and subtracting from (2) we get,
21 y = 126 
=> y = 6
Putting y = 6 in (1), we get: x= 2
=> x= 2, y= 6

Q 4-217x + 131y = 913および131x + 217 y = 827が与えられた場合、x、yは次のようになります。

A -1,6

B -3,2

C -12,13

D -16,18

Answer - B

Explanation

217x +131y= 913  ...(a)
131 x+ 217 y= 827 ...(b)

It is a special case in which coefficients of x and y in (a) are interchanged in (b)
Adding (a) and (b) , we get : 348(x+y)= 1740 => x+y = 5 ...(a)
Subtracting (b) from (a), we get: 86(x-y) = 86 => x-y =1 ...(b)
Adding (a) and (b), we get: x= 3, y= 2

Q 5 -hのどの値に対して、連立方程式hx-y-2 = 0および6x-2y-3 = 0は一意の解を持っていますか？

A -2

B -3

C -4

D -5

Answer - B

Explanation

For, a unique solution, we must have a₁/a
        
          ≠ b
         ₁/b
         ₂ h/6 ≠ -1/-2 => h/6 ≠ 1/2 => h = 3

Q 6 -hのどの値に対して、連立方程式x + 2y + 7 = 0および2x + hy + 14 = 0には、無限の数の解がありますか？

A -3

B -4

C -5

D -6

Answer - B

Explanation

For infinite solutions, we have a₁/a₂ = b₁/b₂= c₁/c₂;
h1/2 = 2/h = 7/14 => h=4.

Q 7-連立方程式hx-10y-3 = 0および3x-5y-7 = 0のhの値について、解はありませんか？

A -6

B -5

C -4

D -3

Answer - A

Explanation

For no solution, we have a₁/a₂ = b₁/b₂ ≠ c₁/c₁
∴ h/3 = -10/-5 ≠-3/-7 => h = 6