Question

4x-5y+16=0 and 2x+y-6=0 solve this ​

Answer 1

Answer:

We have system of equation:

A:4x−5y+16=0−−−(1)

B:2x+y−6=0−−−−(2)

For A

x0−41

y16/504

For B

x031

y604

From solving equation (1) & (2), we get only one point (1,4) as

−7y+28=0⇒y=4

4x−5y+16=04x+2y−12=0−−+

,x=

2

6−y

=

2

2−4

=1

So, the vertices are (−4,0),(1,4) & (3,0)

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Answer 2

Step-by-step explanation:

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$\red{\bold{\underline{\underline{❥Question᎓}}}}$4x-5y+16=0 and 2x+y-6=0

Answer:

⟹$4x - 5y + 16 = 0$

⟹ $x = \frac{16 + 5y}{4}$

put value of x in another equation

⟹ $2( \frac{16 + 5y}{4} ) + y - 6 = 0$

⟹ $\frac{32 + 10y}{4} + y - 6 = 0$

⟹ $32 + 10y + 4y - 24 = 0$

⟹ $14y + 8 = 0$

⟹ $14y = - 8$

⟹ $y = - \frac{8}{14} = - \frac{4}{7} [/</p><p>tex]</p><p>put y value in </p><p>⟹ [tex]x = \frac{32 + 10 \times - \frac{4}</p><p>{7} }{4} = 8 + 10 \times - \frac{4}{7}$

⟹ $18 \times - \frac{4}{7} = - \frac{72}{7}$

Hope it helps you..!!!

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