Question

Please help me!! to solve.
(1) 5x + 2y = 7
6X + 5y = 38
eo find X, y value
(2) 2x + 3y + 2z = 2 , 3x - 2y - 2z= 3 find X, Y, Z value.​

Answer 1

Answer:

part1

5x+2y=7-(1)

6x+5y=38-(2)

from (1)

5x=7-2y

x=(7-2y)/5 -(3)

By putting equation (3) in equation(1), we get

6(7-2y)/5 + 5y=38

(42-12y)/5 + 5y = 38

(42-12y+25y)/5=38

(42+13y)=38×5

42+13y=190

13y=190-42

13y=148

y=148/13

putting value of y in equation (3) in (2), we get

$x = \frac{7 - 2y}{5}$

x=[7-2(148/13)]/5

x=[7-296/13]/5

x=(91-296)/13×5

×=-205/65

x=-41/13

Answer 2

Required solution: x = $-\dfrac{41}{13}$, y = $\dfrac{148}{13}$.

Step-by-step explanation:

The given equations are

5x + 2y = 7 ... ... (1)

6x + 5y = 38 ... ... (2)

From (1), we get

5x = 7 - 2y

⇒ x = $\dfrac{7-2y}{5}$ ... ... (3)

Substituting x = $\dfrac{7-2y}{5}$ in (2), we get

6 × $\dfrac{7-2y}{5}$ + 5y = 38

⇒ $\dfrac{42-12y+25y}{5}$ = 38

⇒ 42 + 13y = 190

⇒ 13y = 148

⇒ y = $\dfrac{148}{13}$

Now putting y = $\dfrac{148}{13}$ in (3), we get

x = $\dfrac{7-2\times\dfrac{148}{13}}{5}$

⇒ x = $\dfrac{\dfrac{91-296}{13}}{5}$

⇒ x = $\dfrac{-205}{65}$

⇒ x = $-\dfrac{41}{13}$

So, the required solution is

x = $-\dfrac{41}{13}$, y = $\dfrac{148}{13}$

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