Question

The value m for which the system of equations 5mx-2y=1 and 10x+y=3 has a unique solutions

Answer 1

Given us the system equations as,

• 5mx - 2y = 1
• 10x + y = 3

Comparing each equation with the standard form of a linear equation in two variable i.e., ax + by = c

We have,

• a₁ = 5m ; b₁ = -2 ; c₁ = 1
• a₂ = 10 ; b₂ = 1 ; c₂ = 3

For a system of equations, there exists unique solutions only when the variables of the equations follow the following condition:

⇒ a₁ / a₂ ≠ b₁ / b₂

⇒ 5m / 10 ≠ -2 / 1

⇒ 5m ≠ -20

⇒ m ≠ -4

So, For any value of m expect -4, the system of equations will have unique solutions which means the lines represented by these two equations will intersect each other only at one point.

Some Information :-

☞ A linear equation is a an algebraic expression of variables. which is of the form ax + by = c

Where, x & y are variables.

☞ For infinite many solutions of a system of equations, the following condition holds true regarding the coefficients of variables and also the constant term(if any) :

⇒ a₁ / a₂ = b₁ / b₂ = c₁ / c₂

Answer 2

Answer :

☞ $\boxed{m ≠ -4}$

Explanation :

Given Equations :

• 5mx - 2y = 1
• 10x + y = 3

According to the question :

These given Equations are in the form,

⇛$\boxed{a1 x + b1 y = 0}$

⇛$\boxed{a2 x + b2 y = 0}$

So, Our Equation will be :

• a1 = 5m , b1 = -2
• a2 = 10 , b2 = ( y ) as 1

For a zero solution, we must have,

➳ $\frac{a1}{a2}≠\frac{b1}{b2}$

➳ $\frac{5m}{10}≠\frac{-2}{1}$

➳ $✯\boxed{cross\:multiply\:✖}✯$

➳ ${5m × 1 }\:≠\:{-2 × 10 }$

➳ $5m\:≠\:{-20}$

➳ $m\:≠\frac{-20}{5}$

➳ $\bold{m\:≠\:-4}$

∴ The given equations will have a unique solution , if m ≠ -4 .

So, It's Done !!