Question

If the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution,
which of the following is true?

Answer 1

Concept:

First order equations include linear equations. In the coordinate system, the linear equations are defined for lines. A linear equation in one variable is one in which there is a homogeneous variable of degree 1 (i.e., only one variable). Multiple variables may be present in a linear equation. Linear equations in two variables, for example, are used when a linear equation contains two variables. For instance, 2x - 3 = 0, 2y = 8, m + 1 = 0, x/2 = 3, x + y = 2, and 3x - y + z = 3 are examples of linear equations.

Given:

the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution

Find:

which of the following is true?

Solution:

Using Cramer's rule,

For a unique solution,

a₁/a₂≠ b₁/b₂

5/8≠k/25

k≠125/8

Therefore, If the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution, k≠125/8

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Answer 2

Answer:

Concept:

A linear equation is an equation in which the highest power of the variable is always 1. It is also known as a one-degree equation. The standard form of a linear equation in one variable is of the form Ax+B=0. Here, x is a variable, A is a coefficient and B is constant. The standard form of a linear equation in two variables is of the form Ax + By = C. Here, x and y are variables, A and B are coefficients and C is a constant.

Find:

If the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution,

which of the following is true?

Given:

If the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution,

which of the following is true?

Step-by-step explanation:

The equation are

5x+ky−3=08

x+25y+k=0

For the system to have unique solution

$\frac{ a_{1} }{ a_{2} } = \frac{ b_{1} }{ b_{2} }$

$\frac{5}{8} = \frac{k}{25}$

$k = \frac{125}{8}$

If the pair of linear equations 5x+ky−3=0 and 8x+25y+k=0 has a unique solution,

$k = \frac{125}{8}$

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