Simple System of Equations Part 3

Algebra Level 4

x y z + 3 x + 3 y z = 106 x y z + 3 z + 3 x y = 79 x y z + 3 y + 3 x z = 82 ( x + 1 ) ( y + 1 ) ( z + 1 ) = ? \begin{aligned}xyz + 3x + 3yz&=&106 \\ xyz + 3z + 3xy&=&79 \\ xyz+ 3y + 3xz&=&82 \\ (x+1)(y+1)(z+1) &=& \ ? \end{aligned}

See Part 1 and Part 2 .


The answer is 90.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

Chew-Seong Cheong
Apr 30, 2015

Expanding ( x + 1 ) ( y + 1 ) ( z + 1 ) = x y z + x y + y z + z x + x + y + z + 1 (x+1)(y+1)(z+1) = xyz + xy + yz + zx + x + y + z +1

We note that adding the three given equations we have:

{ x y z + 3 x + 3 y z = 106 x y z + 3 z + 3 x y = 79 x y z + 3 y + 3 z x = 82 3 ( x y z + x y + y z + z x + x + y + z ) = 267 \begin{cases} xyz + 3x + 3yz = 106 \\ xyz + 3z + 3xy = 79 \\ xyz + 3y + 3zx = 82 \end{cases} \Rightarrow 3(xyz + xy + yz + zx + x + y + z) = 267

( x + 1 ) ( y + 1 ) ( z + 1 ) = 267 3 + 1 = 89 + 1 = 90 \Rightarrow (x+1)(y+1)(z+1) = \dfrac{267}{3} +1 = 89 + 1 = \boxed{90}

Moderator note:

Well done. A common appraoch students make is to first determine the values of x , y , z x,y,z which is not always the case for all system of equations.

Betty BellaItalia
Apr 22, 2017

Panda Đỗ
Apr 30, 2015

we have:A= (x+1)(y+1)(z+1)=xyz+x+y+z+xy+yz+zx+1, so when we plus both 3 equations, we have 3(xyz+x+y+z+xy+yz+zx)=267 <=> 3(A-1)=267, so A=90

Moderator note:

Superb!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...