The solution of the simultaneous linear equations $2 x + y = 6$ and $3 y = 8 + 4 x$ will also be satisfied by which one of the following linear equations?

x + y = 5

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2 x + y = 9

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2 x - 3 y = 10

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2 x + 3 y = 6

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Solution

The correct option is C $x + y = 5$
$3 y = 8 + 4 x$
$\Rightarrow - 4 x + 3 y = 8$ ...(1)

$2 x + y = 6$
$\Rightarrow 4 x + 2 y = 12$ ....(2)
Adding equations (1) and (2), we get
$5 y = 20 \Rightarrow y = 4$
Hence, $x = 1$ .
Only option 1 satisfies the above values.

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Similar questions

Q. Find the solution of the simultaneous linear equations using the graphical method.

3 y - 2 x = 6

2 x + y = 2

Q. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

(i)

x + y = 5, 2 x + 2 y = 10

(ii)

x - y = 8, 3 x - 3 y = 16

(iii)

2 x + y - 6 = 0, 4 x - 2 y - 4 = 0

(iv)

2 x - 2 y - 2 = 0, 4 x - 4 y - 5 = 0

Which of the following pairs of linear equations is inconsistent?
$1. x + 3 y - 5 = 0, 2 x + y = 0$
$2. 2 x + y = 1, 2 y = 2 - 4 x$
$3. 2 x - y = 2, 2 y - 4 x = 2$

Q. Express the following linear equations in the form ax + by + c = 0 and indicate values of a, b and c in each case :

(i) 2 x + 3 y = 9.

(i i) x + \frac{y}{5} + 10 = 0

(i i i) 2 x + 3 y = 6

(i v) x = 3 y

(v) 2 x = 5 y

Q. Solve the following pair of linear (simultaneous) equations by the method of elimination:

2 x + 3 y = 8

2 x = 2 + 3 y

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The solution of the simultaneous linear equations 2x+y=6 and 3y=8+4x will also be satisfied by which one of the following linear equations?

The correct option is C x+y=53y=8+4x⇒−4x+3y=8 ...(1)2x+y=6⇒4x+2y=12 ....(2)Adding equations (1) and (2), we get5y=20⇒y=4Hence, x=1.Only option 1 satisfies the above values.

The solution of the simultaneous linear equations $2 x + y = 6$ and $3 y = 8 + 4 x$ will also be satisfied by which one of the following linear equations?

The correct option is C $x + y = 5$
$3 y = 8 + 4 x$
$\Rightarrow - 4 x + 3 y = 8$ ...(1)

$2 x + y = 6$
$\Rightarrow 4 x + 2 y = 12$ ....(2)
Adding equations (1) and (2), we get
$5 y = 20 \Rightarrow y = 4$
Hence, $x = 1$ .
Only option 1 satisfies the above values.