How do you solve $x+2y+z=34$ , $x+2y+2z=38$ , $2x+y+3z=54$ ?

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Answer 1 · 2016-07-02T04:44:28

$x=18,y=6,z=4$

$color(green)("Three given equations are")$

$x+2y+z=34.......(1)$ ,

$x+2y+2z=38.......(2)$ ,

$2x+y+3z=54......(3)$

$color(blue)("Subtracting (1) from(2) we have")$

$x+2y+2z-(x+2y+z)=38-34$

$=>cancelx+cancel(2y)+2z-cancelx-cancel(2y)-z=4$

$=>z=4$

$color(blue)("Inserting the value of z in (1)")$

$x+2y+4=34$ ,

$x+2y=30.....(4)$

$color(blue)("Inserting the value of z in (3)")$

$2x+y+3xx4=54$

$2x+y =42.......(5)$

$color(blue)("Multiplying (4) by 2 and subtracting the resulting equation from (5)")$

$2x+y -2xx(x+2y)=42-2xx30$

$=>2x+y -2x-4y=42-60$

$=>-3y=-18$

$=>y=6$

$color(blue)("Inserting the value of y in (4) "$

$x+2xx6=30$

$=>x=18$

How do you solve x+2y+z=34x+2y+z=34, x+2y+2z=38x+2y+2z=38, 2x+y+3z=542x+y+3z=54?