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Byju's Answer

Standard IX

Mathematics

Linear Equation in 2 Variables

Which of the ...

Question

Which of the following equations has $x = 1$ and $y = 1$ as their solutions?

2 x + y = 4

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x + y = 2

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x - y = 1

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x + 2 y = 3

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Solution

The correct options are
A $x + y = 2$
C $x + 2 y = 3$
$(A) 2 x + y = 4$
Put the value $x = 1$ and $y = 1$ we get
$2 \times 1 + 1 \times 1 = 2 + 1 = 3 \neq 4$
Then $x = 1$ and $y = 1$ is not the solution
$(B) x + y = 2$
Put the value $x = 1$ and $y = 1$ we get
$1 \times 1 + 1 \times 1 = 2 = 4$
It satisfies 2
Then $x = 1$ and $y = 1$ is the solution
$(C) x - y = 1$
$1 \times 1 - 1 \times 1 = 1$
$1 - 1 = 0 \neq 1$
Not satisfies
Then $x = 1$ and $y = 1$ is not the solution
$(D) x + 2 y = 3$
Put the value $x = 1$ and $y = 1$ we get
$1 \times 1 + 2 \times 1 = 1 + 2 = 3 = 3$
Then $x = 1$ and $y = 1$ is the solution
Thus, equation in (B) and (D) has their solution $x = 1$ and $y = 1$

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Similar questions

Q. Question 10
A pair of linear equations which has a unique solution x = 2 and y = - 3 is
(A) x + y = 1 and 2x – 3y = -5
(B) 2x + 5y = - 11 and 4x + 10y = - 22
(C) 2x – y = 1 and 3x + 2y = 0
(D) x – 4y + 14 = 0 and 5x – y – 13 = 0

Q. What is the solution for the following pair of linear equations?

\frac{1}{2 x} - \frac{1}{y} = - 1

\frac{1}{x} + \frac{1}{2 y} = 8

(Given:

x \neq 0 a n d y \neq 0)

What is the solution for the following pair of linear equations?

$\frac{1}{2 x} - \frac{1}{y} = - 1$

$\frac{1}{x} + \frac{1}{2 y} = 8$

(Where $x \neq 0, y \neq 0)$

Q. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)

\frac{1}{2 x} + \frac{1}{3 y} = 2; \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

(ii)

\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2; \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1

(iii)

\frac{4}{x} + 3 y = 14; \frac{3}{x} - 4 y = 23

(iv)

\frac{5}{x - 1} + \frac{1}{y - 2} = 2; \frac{6}{x - 1} - \frac{3}{y - 2} = 1

(v)

\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15

(vi)

6 x + 3 y = 6 x y; 2 x + 4 y = 5 x y

(vii)

\frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = - 2

(viii)

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}; \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}

Q. What is the solution for the following pair of linear equations?

\frac{1}{2 x} - \frac{1}{y} = - 1

\frac{1}{x} + \frac{1}{2 y} = 8

(Where

x \neq 0, y \neq 0)

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Which of the following equations has x=1 and y=1 as their solutions?

Which of the following equations has $x = 1$ and $y = 1$ as their solutions?