Question

Mam dwa proste układy równań do rozwiązania!

1.
{ 10(y+2)+3(x-5)= x-5
{ 25(y+2)+3(x-5)= -10

2.
{ 2(2x-y)+3(x+3y)= 42
{ 5(x+3y)-8(2x-y) - 32

Answer 1

1.
{ 10(y+2)+3(x-5)= x-5
{ 25(y+2)+3(x-5)= -10

{ 10y+20+3x-15= x-5
{ 25y+50+3x-15= -10

{ 10y +2x = -10 z tego wyznaczamy x więc 2x = -10y-10 / 2
{ 25y + 3x= -45 x = -5y-5

{x = -5y-5
{25y + 3(-5y-5) = -45

{x = -5y-5
{25y -15y-15 = -45

{x = -5y-5
{10y= -30 / 10

{x = -5y-5
(y= -3

{x = -5(-3) -5
{y=-3

{x = -15-5
{y=-3

{x= -20
{y=-3

2. rozwiązałam robiąc w drugim równaniu = 32 zamiast -32
jeżeli powinno być -32 to pisz to poprawię

{ 2(2x-y)+3(x+3y)= 42
{ 5(x+3y)-8(2x-y) = -32

{ 4x-2y+3x+9y= 42
{ 5x+15y -16x+8y =-32

{ 7x+7y = 42 /7
{-11x+ 23y = -32

{ x+y = 6 z tego wyznaczamy x więc x= 6-y
{-11x+ 23y = -32

{ x= 6-y
{-11(6-y)+ 23y = -32

{ x= 6-y
{-66 + 11y+ 23y = -32

{ x= 6-y
{34y = 34 / 34

{ x= 6-y
{y = 1

{ x= 6-1
{y = 1

{ x= 5
{y = 1