Question

Solve the simultaneous equations
5x + 3y - 41
2x + 3y - 20

Answer 1

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.