How do you solve the system of equations: $x+y+z = 105$ , $10y-z = 11$ , $2x-3y = 7$ ?

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Answer 1 · 2016-09-27T07:21:18

$x=17$ , $y=9$ and $z=79$

Given:

$x+y+z = 105$

$10y-z = 11$

$2x-3y = 7$

Add the first two equations together to eliminate $z$ and get:

$x+11y = 116$

Multiply this equation by $2$ to get:

$2x+22y = 232$

Subtract the original third equation to eliminate $x$ and get:

$25y = 225$

Divide both sides by $25$ to get:

$color(blue)(y = 9)$

Substitute this value of $y$ into the third equation to get:

$2x-27 = 7$

Add $27$ to both sides to get:

$2x = 34$

Divide both sides by $2$ to get:

$color(blue)(x = 17)$

Substitute the values for $x$ and $y$ into the first equation to get:

$17+9+z = 105$

Subtract $26$ from both sides to get:

$color(blue)(z = 79)$

How do you solve the system of equations: x+y+z = 105x+y+z = 105, 10y-z = 1110y-z = 11, 2x-3y = 72x-3y = 7 ?