Question

select the correct systems of equations.
which systems of linear equations have no solution?
$x + y + z = 1,100$
$x - 2y - z = -500$
$2x + 3y + 2z = 2,600$

$x + y + z = 1,500$
$x - y - z = -500$
$2x + y + z = 2,000$

$x + y + z = 1,400$
$x - 2y - z = -500$
$2x + 2y + 2z = 2,700$

$x + y + z = 1,400$
$-0.5x - 0.5y - 0.5z = -900$
$2x + 3y + 2z = 3,000$

$x + y + z = 1,900$
$x - y - 2z = -2,000$
$2x + 2y + z = 1,100$

$x + y + z = 2,400$
$2x - 2y + 2z = 700$
$x + 3y + z = 2,400$

Explanation:

Step1: Recall no-solution condition

A system of linear equations has no solution if the equations are inconsistent (they represent parallel planes with no intersection, or lead to a contradiction like $0=\text{non-zero}$). We can test by combining equations to eliminate variables.

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System 1:

$x+y+z=1,100$
$x-2y-z=-500$
$2x+3y+2z=2,600$

Step1: Add first 2 equations

$(x+y+z)+(x-2y-z)=1100+(-500)$
$2x-y=600$

Step2: Multiply first eq by 2

$2x+2y+2z=2200$

Step3: Subtract from third eq

$(2x+3y+2z)-(2x+2y+2z)=2600-2200$
$y=400$

Step4: Solve for x

$2x-400=600 \implies x=500$

Step5: Solve for z

$500+400+z=1100 \implies z=200$
This system has a solution.

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System 2:

$x+y+z=1,500$
$x-y-z=-500$
$2x+y+z=2,000$

Step1: Add first 2 equations

$(x+y+z)+(x-y-z)=1500+(-500)$
$2x=1000 \implies x=500$

Step2: Substitute x into first eq

$500+y+z=1500 \implies y+z=1000$

Step3: Substitute x into third eq

$2(500)+y+z=2000 \implies 1000+y+z=2000 \implies y+z=1000$
Consistent, has infinitely many solutions.

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System 3:

$x+y+z=1,400$
$x-2y-z=-500$
$2x+2y+2z=2,700$

Step1: Multiply first eq by 2

$2x+2y+2z=2800$

Step2: Compare to third eq

Third eq is $2x+2y+2z=2700$, so $2800=2700$, a contradiction.
This system has no solution.

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System 4:

$x+y+z=1,400$
$-0.5x-0.5y-0.5z=-900$
$2x+3y+2z=3,000$

Step1: Multiply second eq by -2

$x+y+z=1800$

Step2: Compare to first eq

First eq is $x+y+z=1400$, so $1400=1800$, a contradiction.
This system has no solution.

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System 5:

$x+y+z=1,900$
$x-y-2z=-2,000$
$2x+2y+z=1,100$

Step1: Subtract second eq from first

$(x+y+z)-(x-y-2z)=1900-(-2000)$
$2y+3z=3900$

Step2: Multiply first eq by 2

$2x+2y+2z=3800$

Step3: Subtract third eq

$(2x+2y+2z)-(2x+2y+z)=3800-1100$
$z=2700$

Step4: Solve for y

$2y+3(2700)=3900 \implies 2y=3900-8100=-4200 \implies y=-2100$

Step5: Solve for x

$x+(-2100)+2700=1900 \implies x=1300$
This system has a solution.

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System 6:

$x+y+z=2,400$
$2x-2y+2z=700$
$x+3y+z=2,400$

Step1: Subtract first eq from third

$(x+3y+z)-(x+y+z)=2400-2400$
$2y=0 \implies y=0$

Step2: Simplify second eq

$2x+2z=700 \implies x+z=350$

Step3: Substitute into first eq

$x+0+z=2400 \implies x+z=2400$

Step4: Check consistency

$350=2400$, a contradiction.
This system has no solution.

Answer:

The systems with no solution are:

1. $x+y+z=1,400$; $x-2y-z=-500$; $2x+2y+2z=2,700$
2. $x+y+z=1,400$; $-0.5x-0.5y-0.5z=-900$; $2x+3y+2z=3,000$
3. $x+y+z=2,400$; $2x-2y+2z=700$; $x+3y+z=2,400$