Question

5.
Solve the equations:
x + 3y = 13
x2 + 3y2 = 43​

Answer 1

Resposta:

see below

Explicação passo-a-passo:

hi come on, in this system of equations we will use the substitution method to find the solutions, note:

$x+3y = 13\Rightarrow x = 13-3y$, we know that ...

$x^2 + 3y^2 = 43\Rightarrow (13-3y)^2 + 3y^2 = 43\Rightarrow 12y^2-78y+126=0\Rightarrow$

$2y^2-13y+21=0\Rightarrow y^2-\frac{13y}{2} +\frac{21}{2}=0\Rightarrow (y-\frac{13}{4})^2 - \frac{169}{16}+\frac{21}{2 }=0\Rightarrow$

$(y-\frac{13}{4})^2 = \frac{1}{16}\Rightarrow y-\frac{13}{4}=\pm\frac{1}{4}\Rightarrow y_1 = \frac{7}{2} \ or \ y_2 = 3$  , now we find the values ​​of x

$x = 13-3y\Rightarrow x_1 = 13 - 3\cdot \frac{7}{2}\Rightarrow x_1 = \frac{5}{2}$

$x= 13-3y\Rightarrow x_2 = 13-3(3)\Rightarrow x_2 = 13-9\Rightarrow x_2 = 4$

hugs.