Question

(3x+2(1-y) = 20
(2(2x+1)-3(2y + 1) = 7
4. Rezolvaţi următoarele sisteme de ecuaţii prin metoda reducerii:
b)
(2x-y=-1
1-2x-y=-1'
c)
fx-2y=-15
1x=y=-5
(2x-3y=-10
12x + 5y = 26
10x+3y=24
-2x+17y=-30
9x-15y=12
a) { x + y = 2
[x=y=4'
(d)
a)
(5x+2y=18
3x - y = 2
(8x+5y=10
20x+9y=18
3x-2y=-10
(18x+25y=-7
9x+19y=-10
(0)
5. Rezolvaţi următoarele sisteme de ecuaţii, prin metoda reducerii:
X
x 3y
4 6
4
2
2x
3
(0,2x-0,3y=0,7
;
1,2x-1,4y=1
e)
(o (2)
h)
b)
y=0
-y=-
f)
5
3
=-3
;
2x_Y = -8
3
6

Answer 1

Explicație pas cu pas:

metoda reducerii:

$\left \{ {{3x+2(1-y) = 20} \atop {2(2x+1)-3(2y + 1) = 7}} \right.$

$\left \{ {{3x + 2 - 2y = 20} \atop {4x + 2 - 6y - 3 = 7}} \right.$

$\left \{ {{3x - 2y = 18} | \cdot ( - 3)| \atop {4x - 6y = 8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \right.$

$\left \{ {{ - 9x + 6y = - 54} \atop {4x - 6y = 8} \: \: } \right.$

------

$- 5x = - 46 = > \bf x = \frac{46}{5}$

$4 \cdot \frac{46}{5} - 6y = 8 \\ 6y = \frac{184}{5} - 8 \\ 6y = \frac{144}{5} = > \bf y = \frac{24}{5}$

b)

$\left \{ {{2x-y = -1} \atop {1-2x-y = -1}} \right.$

$\left \{ {{2x-y = -1} \atop {-2x-y = -2}} \right.$

----

$-2y = -3 = > \bf y = \frac{3}{2}$

$2x - \frac{3}{2} = -1 <=> 2x = \frac{1}{2} = > \bf x = \frac{1}{4}$