Respostas
Explicação passo-a-passo:
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 = 13-3yx+3y=13⇒x=13−3y , we know that ...
x^2 + 3y^2 = 43\Rightarrow (13-3y)^2 + 3y^2 = 43\Rightarrow 12y^2-78y+126=0\Rightarrowx
2
+3y
2
=43⇒(13−3y)
2
+3y
2
=43⇒12y
2
−78y+126=0⇒
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\Rightarrow2y
2
−13y+21=0⇒y
2
−
2
13y
+
2
21
=0⇒(y−
4
13
)
2
−
16
169
+
2
21
=0⇒
(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(y−
4
13
)
2
=
16
1
⇒y−
4
13
=±
4
1
⇒y
1
=
2
7
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⇒x
1
=13−3⋅
2
7
⇒x
1
=
2
5
x= 13-3y\Rightarrow x_2 = 13-3(3)\Rightarrow x_2 = 13-9\Rightarrow x_2 = 4x=13−3y⇒x
2
=13−3(3)⇒x
2
=13−9⇒x
2
=4
hugs.