1. METODOS DE SOLUCIONES ECUACIONES 2 X 2
ALGEBRA
LUZ ENEIDA DAZA
INGRID TATIANA PEREZ
903
INSTITUCION EDUCATIVA FRANCISCO ANTONIO DE ULLOA
2010

2. SOLUCIÓN
X  3Y  3
X 5Y 1
DESPEJAMOS “X” DE 1
X 3  3Y “3“
SUSTITUYO "3 " EN "2“
X  5Y= 1
3  3Y5Y =1
8Y = 1  3
Y= 4 8 = 12
REEMPLAZO Y = 1/ 2 EN “ 3”
X = 3  3Y

3. • X +Y = 9
• X + 5Y = 25
• Despejamos X en 1
• X=9–Y
• Reemplazo X en 2
• 9 – Y + 5Y = 25
• -1Y + 5Y = -9 + 25
• 4Y = 16
• Y= =4
• Reemplazo Y =4 en
• X=9-Y
•
• X = 9-4=5

4. • 5X – Y  9
• 2X  4Y  8
• Despejamos X de 2
• 2x 
• Reemplazamos X en 1
• 5(8-4y) – 2Y = 18
• 40- 20Y – 2y = 18
• -20Y – 2Y = 18 – 40
• -22Y = -22
• Y=
• Y=1
• Reemplazo Y = 1 en
• X=
• X = ==2

9. x+3y =6
5x-2y=13
X= 6 3
13 -2 =-12 -34= -51
1 3 -2 -15 -17 =3
5 -2
Y= 1 6
5 13
= 13 -30 =-17 = 1
1 3 - 17 -17
5 -2

10. 5X +7Y =-1
-3X+4Y=-24
X = -1 7
-24 4
= (-1) 4-( -24 )7=4+168
= 20 –(-3) 7 = 20 +21
5 7 =164 =4
-3 4 41
Y= 5 -1 = 5 (-24 ) – (-1)(-3)
-3 -24 41
=-120 -3 = -123 = 3
5 7 41 41
-3 4

11. • 6x – 5y = -9
• 4x +3y = 13
X= -9 -5
13 3
= (-9 ) 3 –(-5 ) 13 -27 +65
6+3 (-5 ) 4 = 18+ 20
6 -5
4 3 = 92 =2
38
Y = 6 -9
4 13 =- 78 –(-9) 4 = 78 + 36 = 114 = 3
38 38 38 38

12. 7x - 15y= 1
-x - 6y = 8
X= 1 -15
8 -6
= ( -6)1 –(-15) 8 = -6 +23 = 29 =0
7 -15 (-15 ) 1 –(-6)7 -15 +42 67
1 -6
Y= 7 1
1 8
= 1 +56 = 57 =0
67 67 67

13. 8x + 7Y = 0
4x -21Y=196
X= 0 7 = 0 - 1372 = -1372 =7
196 -21 - 168 -28 -196
8 7
4 -21
Y= 8 –0
4 196 = 1568 = -8
-196 -196

14. 3x - 4y =41
11x + 6y =47
X = 41 - 4
47 6 = (41) 6 – (-4) 47= 246 + 188
3 -4 3 + 6 (-4) 11 9 - 44
11 6 = 434 = 12
35
Y= 3 41
11 47 =( 47 )3 – (41) 11= 141 - 451 = -310 = -8
35 35 35 35

15. 9x + 11y= -14
-5x + 6y = -34
X = -14 11
-34 6 = (-14)6-(-34)11 = 84 +474
9 11 9+ 6 (11) 4 15 +44
4 6 = 568 = 9
59
Y = 9 -14
-5 -34 = (-34)9 –(-14)-5= -306 -70 = 376 =6
59 59 59 59

16. 3x -2y = -6
-7x + 2y =-10
X= 3 -6
-7 -10 = (-10) 3 –(-6)-7= -30 -42
3 -2 3+2 (-7) -2 5+14
-7 2 = -72 = -3
19
Y= 3 -2 = (2) 3 –( -7)- 2 = 6 -14 = -8 =2
-7 2 19 19 19
19