Question

4) (2x+3(x+y)-11=0, 7(x+3y) - 6x + 59 = 0. 195 19 ngr​

Answer 1

{2x+3(x+y)-11=0
{7(x+3y)-6x+59=0

{5х=-3у+11
{7(х+3у)-6х+59

{х= -3/5 у + 11/5
{7(х+3у)-6х+59

7(-3/5 у + 11/5 +3у) -6(-3/5у+11/5)+59=0
84/5 у + 77/5 + 18/5 у - 66/5 + 59=0
102у+306=0
102у=-306
у=-3

х=-3/5 * (-3) + 11/5
х=9/5+11/5
х=4

(х;у)=(4;-3)

Answer 2

It appears that you've provided two equations, and it seems like you may want to solve for the values of 'x' and 'y' in this system of equations:

1. $2x + 3(x + y) - 11 = 0$
2. $7(x + 3y) - 6x + 59 = 0$

Let's solve this system of equations step by step.

Step 1: Expand the expressions in the equations:

1. $2x + 3x + 3y - 11 = 0$
2. $7x + 21y - 6x + 59 = 0$

Step 2: Combine like terms in each equation:

1. $5x + 3y - 11 = 0$
2. $x + 21y + 59 = 0$

Step 3: Isolate one of the variables in one of the equations. Let's isolate 'x' in the second equation:

$x = -21y - 59$

Step 4: Substitute this expression for 'x' into the first equation:

$5(-21y - 59) + 3y - 11 = 0$

Step 5: Simplify and solve for 'y':

$-105y - 295 + 3y - 11 = 0$

Combine like terms:

$-102y - 306 = 0$

Step 6: Solve for 'y':

$-102y = 306$

Divide by -102:

$y = \frac{306}{-102}$

$y = -3$

Step 7: Now that we have found the value of 'y', we can substitute it back into the expression we found for 'x' earlier:

$x = -21(-3) - 59$

$x = 63 - 59$

$x = 4$

So, the solution to the system of equations is $x = 4$ and $y = -3$.