Question

3x+4y+z,=11, 12x+8y+z=63, 9x+6y+z=64​

Answer 1

Answer:

⇒ $x = 18$

⇒ $y = -27.5$

⇒ $z = 67$

Step-by-step explanation:

Given :

To solve for x , y & z,

If,

$3x+4y+z=11$

$12x+8y+z=63$

$9x+6y+z=64$

Solution :

By using elimination method (finding the value of variables by eliminating other variables),.

$3x+4y+z=11$ ...(i)

$12x+8y+z=63$ ...(ii)

$9x+6y+z=64$ ...(iii)

By subtracting (i) from (iii) (eliminating z)

⇒ $(9x+6y+z)-(3x+4y+z)=64-11$

⇒ $9x - 3x + 6y - 4y +z -z=53$

⇒ $6x + 2y=53$ ...(iv)

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By subtracting (i) from (ii) (eliminating z)

⇒ $(12x+8y+z) - (3x+4y+z)=63 - 11$

⇒ $12x- 3x + 8y - 4y +z-z=52$

⇒ $9x + 4y=52$ ...(v)

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By subtracting 2 × (v) from 4 × (iv)

⇒ $4 \times (6x + 2y) - 2 \times(9x + 4y)= 4 \times 53-2 \times 52$

⇒ $(24x + 8y) - (18x +8y)= 212-104$

⇒ $24x - 18x + 8y - 8y= 108$

⇒ $6x= 108$

⇒ $x= \frac{108}{6} = 18$

⇒ $x=18$

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By substituting value of x in (iv)

⇒ $6x + 2y=53$

⇒ $6(18) + 2y=53$

⇒ $108 + 2y=53$

⇒ $2y=53-108$

⇒ $2y=-55$

⇒ $y=-\frac{55}{2} = -27.5$

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By substituting value of x & y in (i)

⇒ $3x+4y+z=11$

⇒ $3(18)+4(-27.5)+z=11$

⇒ $54-110+z=11$

⇒ $-56+z=11$

⇒ $z=11+56$

⇒ $z=67$

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