Question

1) - За – 5 (4а – 3b) + 2 (3b - а - 1), если а=2\9, b=3\7 2) 6х – 3 (2y - (у - (1 - 4у))), если 2х + Зу = 5.​

Answer 1

Ответ:

1) - За – 5 (4а – 3b) + 2 (3b - а - 1), раскрываем скобки

-3a-20a+15b+6b-2a-2,

-25a+21b-2, подставляем значения a и b

-25*2/9+21*3/7-2=13/9

2) 6х – 3 (2y - (у - (1 - 4у))), раскрываем скобки

3(2x-(2y-(5y-1))),

3(2x-(2y-5y+1),

3(2x-(-3y+1),

3(2x+3y-1), подставляем значение 2x+3y

3(5-1)=3*4=12.

Answer 2

To evaluate the given expressions, let's substitute the given values of 'a' and 'b' and solve the equations.

1. $-5(4a - 3b) + 2(3b - a - 1)$ with $a = \frac{2}{9}$ and $b = \frac{3}{7}$:

First, replace 'a' and 'b' with their values:

$-5\left(4 \left(\frac{2}{9}\right) - 3\left(\frac{3}{7}\right)\right) + 2\left(3\left(\frac{3}{7}\right) - \frac{2}{9} - 1\right)$

Now, perform the arithmetic operations step by step:

$-5\left(\frac{8}{9} - \frac{9}{7}\right) + 2\left(\frac{9}{7} - \frac{2}{9} - 1\right)$

$-5\left(\frac{56}{63} - \frac{81}{63}\right) + 2\left(\frac{81}{63} - \frac{14}{63}\right)$

$-5\left(\frac{-25}{63}\right) + 2\left(\frac{67}{63}\right)$

$\frac{-5 \cdot (-25)}{63} + \frac{2 \cdot 67}{63}$

$\frac{125}{63} + \frac{134}{63}$

Now, combine the fractions:

$\frac{125 + 134}{63}$

$\frac{259}{63}$

So, the value of the expression is $\frac{259}{63}$.

2. $6x - 3(2y - (y - (1 - 4y)))$ with $2x + 3y = 5$:

First, simplify the expression inside the parentheses:

$6x - 3(2y - (y - (1 - 4y))) = 6x - 3(2y - (y - 1 + 4y))$

$= 6x - 3(2y - 5y + 1) = 6x - 3(-3y + 1)$

Now distribute the -3:

$= 6x + 9y - 3$

Now, we are also given the condition $2x + 3y = 5$. Let's use this information to find the values of 'x' and 'y':

$2x + 3y = 5$

$2x = 5 - 3y$

$x = \frac{5 - 3y}{2}$

Now, substitute this expression for 'x' into the simplified expression:

$6x + 9y - 3 = 6\left(\frac{5 - 3y}{2}\right) + 9y - 3$

Now, distribute the 6:

$= \frac{6(5 - 3y)}{2} + 9y - 3$

$= \frac{30 - 18y}{2} + 9y - 3$

$= 15 - 9y + 9y - 3$