Question

Tarea 2
Usando el método de sustitución resuelve los siguientes sistemas de ecuaciones lineales de 2x2.

1.
2x − 3y = 7
x+ 5y = 23

2.
6x + 5y = 227
5x + 4y = 188

3.
5x − 2y = 1
3x+ 5y = 13​

Doy corona al que responda con procedimiento.

Answer 1

1.

2x - 3y =7

x + 5y = 23

2x - 3y =7

x = 23 - 5y

2 (23 - 5y) -3y = 7

y = 3

x = 23 - 5 x 3

x = 8

(x.y) = (8,3)

2 x 8 - 3 x 3 =7

8 + 5 x 3 = - 23

7=7

23 = 23

(x,y) = (8,3)

________________

2.

6x + 5y = 227

5x + 4y = 188

$\\ 6x + 5y = 227 \\ x = \frac{188}{5} - \frac{4}{5} y$

$\\ 6( \frac{188}{5} - \frac{4}{5y} ) + 5y = 227$

y = 7

$\\ x = \frac{188}{5} - \frac{4}{5} \times 7$

x = 32

(x,y) = (32,7)

6 x 32 + 5 × 7 = -227

5 x 32 + 4 x 7 = 188

227 = 227

188 = 188

(x,y) = (32,7)

________________

3.

5x − 2y = 1

3x + 5y = 13

$\\ x = \frac{1}{5} + \frac{2}{5} y \\ 3x + 5y = 13$

$\\ 3( \frac{1}{5} + \frac{2}{5} y) + 5y = 13$

y = 2

$\\ x = \frac{1}{5} + \frac{2}{5} \times 2$

x = 1

(x,y) = (1,2)

5 x 1 - 2 x 2 = 1

3 x 1 + 5 x 2 = 13

1 = 1

13 = 13

(x,y) = (1,2)

✓

Answer 2

1. 2x-3y=7

x+5y=23

2(23-5y)-3y=7

y=3

x=23-5×3

x=8

(x,y)=(8;3)

2. 6x+5y=227

5x+4y=188

6(188/5-4/5y)+5y=227

y=7

x=188/5-4/5×7

x=32

(x,y)=(32;7)

3. 5x-2y=1

3x+5y=13

3(1/5+2/5y)+5y=13

y=2

x=1/5+2/5×2

x=1

(x,y)=(1;2)