BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Equations

Expert replies
by nehakhas1 » Fri Jan 30, 2009 4:15 am
x + 2y + z = 8;
2x + y + z = 7
What are the values of x, y, z?
I. x, y, z are positive integers. II. x, y, z are distinct numbers

We can find the ans without these given conditions as well .Then should our answer be E?
Join the discussion
Source: — Data Sufficiency |

Just trying

by Alara533 » Fri Jan 30, 2009 9:48 am
x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
Join the discussion

Re: Just trying

by ajmoney09 » Fri Jan 30, 2009 12:24 pm
Alara533 wrote:x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
I dont agree....

You nee three distinct equations to solve for the three different values.. you are only given two.

What is the OA?
Join the discussion

by Alara533 » Fri Jan 30, 2009 2:38 pm
We need three equations to solve three unknowns when we don't have any other information about to the unknown. But in this case we have additional information.

For eg:- x + y + z = 3 can be solved with only this one equation, assuming we know x, y and z are all positive integers.
Join the discussion

Re: Just trying

by coffee5251 » Fri Jan 30, 2009 3:37 pm
Alara533 wrote:x + 2y + z = 8
2x + y + z = 7

These two equations tell us that y-x = 1

Substituting for y in the first eq...we have
x + 2(1+x) + z = 8
3x + z = 6 ------ (A)

Now from (I), we have x,y and z are positive integers.

Since x and z are positive, z will be > 0.
For equation A to be true X cannot have a value > 1, (if x is > 1 then z will have to be <0)

x = 1 implies, z = 3 and y = 2. and these values solve the first two equations.

So we can find the values from the first statement itself.
Why can't x = 2 and z = 0?
Join the discussion

by masuarezdl » Fri Jan 30, 2009 3:47 pm
Alara533 wrote:We need three equations to solve three unknowns when we don't have any other information about to the unknown. But in this case we have additional information.

For eg:- x + y + z = 3 can be solved with only this one equation, assuming we know x, y and z are all positive integers.
AJmoney09, the thing here is to substract one equation from the other, that way you will be left with two unknowns and only one set of numbers does the job. Let me try to illustrate it:

We subsract equation 2 from equation 1...
x + 2y + z = 8
- 2x - y - z = -7

The remaining equation is -x + y = 1

Then we find the value of y = x + 1, and substitute in any of the two equations given:

x + 2(x+1) + z = 8
x + 2x + 2 + z = 8
3x + z = 6

In this last equation, if all numbers are required to be positive, then x=1 and z=3, only those numbers satisfy the condition. Afterwards, you will only have to substitute to find the value of y.

Coffee5251, the number 0 is neither positive nor negative. Statement 1 indicates that x, y and z are all positive numbers.

Hope it helps.
Join the discussion