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

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Answer 1 · 2016-09-27T07:21:18

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

Given:

$x + y + z = 105$ $x+y+z = 105$

$10 y - z = 11$ $10y-z = 11$

$2 x - 3 y = 7$ $2x-3y = 7$

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

$x + 11 y = 116$ $x+11y = 116$

Multiply this equation by $2$ $2$ to get:

$2 x + 22 y = 232$ $2x+22y = 232$

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

$25 y = 225$ $25y = 225$

Divide both sides by $25$ $25$ to get:

$y = 9$ $color(blue)(y = 9)$

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

$2 x - 27 = 7$ $2x-27 = 7$

Add $27$ $27$ to both sides to get:

$2 x = 34$ $2x = 34$

Divide both sides by $2$ $2$ to get:

$x = 17$ $color(blue)(x = 17)$

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

$17 + 9 + z = 105$ $17+9+z = 105$

Subtract $26$ $26$ from both sides to get:

$z = 79$ $color(blue)(z = 79)$

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 ?