Question

Solve the simultaneous equation

x+2y=5
x²-y²=-3​

Answer 1

(1, 2)

(-13/3, 14/3).

Explanation:

x + 2y = 5

x² - y² = -3​

From the first equation x = 5 - 2y

Substitute for y in the second equation:

(5 - 2y)^2 - y^2 = -3

25 + 4y^2 - 20y - y^2 = -3

3y^2 - 20y + 28 = 0

Factoring:-

(3y - 14(y - 2) = 0

y = 2, 14/3.

So when y = 2, x = 5 - 2(2) = 1.

When y = 14/3, x = 5 - 2(14/3) = -13/3.

Checking the results in the second equation:

(1, 2):-

1^2 - 2^2 = -3 Correct.

(-13/3, 14/3):-

(-13/3)^2 - (14/3)^2 = -3 Correct.

Answer 2

y = 14/3 and 2

x = -13/3 and 1

(-13/3, 14/3) and (1, 2)

Explanation:

`x + 2y = 5 - - - (a) \\ {x}^(2) - {y}^(2) = - 3 - - - (b)`

for equation (a)

make x the subject of the formular:

`x = 5 - 2y - - - (c)`

for equation (b)

`{x}^(2) - {y}^(2) = - 3`

substitute for x as 5 - 2y in equation (b):

`{(5 - 2y)}^(2) - {y}^(2) = - 3 \\ (25 - 20y + 4 {y}^(2) ) - {y}^(2) = - 3 \\ 25 - 20y + 3 {y}^(2) = - 3 \\ {3y}^(2) - 20y + 28 = 0 \\ (3y - 14)(y - 2) = 0`

therefore,

`y = (14)/(3) \: \: and \: \: 2`

substitute for all values of y in equation (c):

for y = 14/3:

`x = 5 - 2( (14)/(3) ) \\ \\ x = 5 - (28)/(3) \\ \\ x = - (13)/(3)`

for y = 2:

`x = 5 - 2(2) \\ x = 5 - 4 \\ x = 1`