Question

if 3x+4y=64 and 2x+3y=56, then
(a)x>y
(b)x<y
(c)x>_y
(d)x<_y

Answer 1

Solution is shown in the attachment.....the final answer is that X<Y
Hope you find it useful
Thank you!

Answer 2

Answer:

Required answer is $x < y$

So, option b is the correct answer.

Step-by-step explanation:

Given two equations are

$3x + 4y = 64....(1)$ and $2x + 3y = 56....(2)$

From equation (1),

$3x + 4y = 64 \\ 3x = 64 - 4y \\ x = \frac{64 - 4y}{3} .....(3)$

and

$2x + 3y = 56 \\ 2x = 56 - 3y \\ x = \frac{56 - 3y}{2} ....(4)$

From equation (3) and (4),

$\frac{64 - 4y}{3} = \frac{56 - 3y}{2} \\ 128 - 8y = 168 - 9y \\ 9y - 8y = 168 - 128 \\ y = 40$

From equation (3),

$x = \frac{64 - 4 \times 40}{3} \\ x = - \frac{96}{3} \\ x = - 32$

Here value of x is (-32) and value of y is 40

So,we can say $x < y$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549

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