Question

The system of linear equations
4x+2y=7, 2x+y=6 has
(A) a unique solution
(B) no solution
(C) an infinite number of solutions
(D) exactly two distinct solutions

Answer 1

Answer:

B) no solution

Step-by-step explanation:

4x+2y=7   -------(1)                    in these two eqns : a1=4,a2=2   &&  b1=2,b2=1

2x+y=6        -------(2)

          a1/a2  =  b1/b2   -------(3)

          4/2=2/1

          2/1=2/1

     The equations satisfy the equation (3)

     so there will be no solution

Note: if it does not statisfy the equation (3)(a1/a2!=b1/b2) then it will have a unique solution and is consistant

Answer 2

Given:

The system of linear equations

4x+2y=7, 2x+y=6

To Find:

The number of solution

Solution:

Two equation has a unique solution when: $\frac{a_{1} }{a_{2} }$ ≠ $\frac{b_{1} }{b_{2} }$

Two equation has no solution when: $\frac{a_{1} }{a_{2} }$ = $\frac{b_{1} }{b_{2} }$ ≠ $\frac{c_{1} }{c_{2} }$

Two equation has infinite number of solutions when: $\frac{a_{1} }{a_{2} }$ = $\frac{b_{1} }{b_{2} }$ = $\frac{c_{1} }{c_{2} }$

According to the question,

4x+2y-7=0 -------(i)                  

2x+y-6=0    -------(ii)

In these two equations :

a1=4             a2=2

b1=2             b2=1

c1 = -7          c2=-6

So,

a1/a2 = 4/2

          =2

b1/b2 = 2/1

         =2  

c1/c2 = -7/-6

∴ $\frac{a_{1} }{a_{2} }$ = $\frac{b_{1} }{b_{2} }$ ≠ $\frac{c_{1} }{c_{2} }$

so there will be no solution.

Hence, The system of linear equations

4x+2y=7, 2x+y=6 has no solution. Option(B)