Pre-Calculus · 7.3 Part 1

3-Variable
Linear Systems

Eliminate one variable at a time — "punch one, then punch another" — and the whole system falls.

0 / 20 solved

⚡ Quick Memory Keys

Step 1

PICK & ELIMINATE
Choose one variable. Use two pairs of equations to cancel it → you get 2 equations, 2 unknowns.

Step 2

SOLVE THE 2×2
Eliminate again → one unknown. Solve it.

Step 3

BACK-SUBSTITUTE
Plug back up: get second, then third variable. Always check in all 3 equations.

Special Cases

0 = 0 → I.M.S. (Infinitely Many Solutions)
0 = k → Inconsistent (No solution)

A
Check If It's a Solution
Q01
Plug & Check

Is \((2, -1, 3)\) a solution to the system?

\(x + y + z = 4\)
\(2x - 3z = -5\)
\(5y + 2z = 1\)
⚡ Fast Method Plug into each equation. If all 3 check ✓ → Yes. If even one fails → No.
Eq①: \(2+(-1)+3=4\) ✓   Eq②: \(2(2)-3(3)=4-9=-5\) ✓   Eq③: \(5(-1)+2(3)=-5+6=1\) ✓
Yes or No?
Q02
Plug & Check

Is \((-1, 0, 4)\) a solution?

\(3x - y + z = 1\)
\(2x - 3z = -14\)
\(5y + 2z = 8\)
⚡ Fast Method Eq①: \(3(-1)-0+4=1\) ✓   Eq②: \(2(-1)-3(4)=-2-12=-14\) ✓   Eq③: \(5(0)+2(4)=8\) ✓
Yes or No?
B
Back Substitution — z is given
Q03
Easy

Solve using back substitution. Write answer as (x, y, z).

\(2x - y + 5z = 16\)
\(y + 2z = 2\)
\(z = 2\)
⚡ Fast Method — Work UP from z z=2 → plug into eq②: \(y+4=2\Rightarrow y=-2\)
→ plug x,y into eq①: \(2x-(-2)+10=16\Rightarrow 2x=4\Rightarrow x=2\)
Answer (x,y,z):
Q04
Easy

Solve using back substitution. Write answer as (x, y, z).

\(4x - 2y + z = 8\)
\(-y + z = 4\)
\(z = 11\)
⚡ Fast Method z=11 → eq②: \(-y+11=4\Rightarrow y=7\) → eq①: \(4x-14+11=8\Rightarrow 4x=11\Rightarrow x=\tfrac{11}{4}\)
Answer (x,y,z):
C
Gaussian Elimination — 3 unknowns
Q05
Medium-Easy

Solve the system. Write answer as (x, y, z).

\(x + y + z = 6\)
\(2x - y + z = 3\)
\(3x - z = 0\)
⚡ Fast Method ①+② → \(3x+2z=9\). Eq③: \(3x=z\). Sub: \(3x+6x=9\Rightarrow 9x=9\Rightarrow x=1\)
z=3, then ①: \(1+y+3=6\Rightarrow y=2\)
Answer (x,y,z):
Q06
Medium-Easy

Solve the system.

\(2x + 2z = 2\)
\(5x + 3y = 4\)
\(3y - 4z = 4\)
⚡ Fast Method Eq①×5: \(10x+10z=10\). Eliminate x with eq②×2: \(10x+6y=8\).
Subtract → \(-6y+10z=2\). Combine with eq③: z=5, then work back: x=-4, y=8.
Answer (x,y,z):
Q07
Medium

Solve the system.

\(4x + y - 3z = 11\)
\(2x - 3y + 2z = 9\)
\(x + y + z = -3\)
⚡ Fast Method Use eq③ to eliminate x from ① and ②. Multiply eq③×4 and subtract from ①, ×2 subtract from ②.
Get 2-eq system in y,z → solve → back-sub.
Answer (x,y,z):
Q08
Medium

Solve the system.

\(x + 2y + z = 1\)
\(x - 2y + 3z = -3\)
\(2x + y + z = -1\)
⚡ Fast Method ①−②: \(4y-2z=4\) → \(2y-z=2\). ①−③×\tfrac{1}{2}\): eliminate x between pairs.
Solve 2×2 → z=0, y=1, x=-1.
Answer (x,y,z):
Q09
Medium

Solve the system.

\(3x - 2y + 4z = 1\)
\(x + y - 2z = 3\)
\(2x - 3y + 6z = 8\)
⚡ Fast Method Multiply eq②×3 subtract from ①: eliminate x → get eq in y,z.
Multiply eq②×2 subtract from ③: another y,z eq. Solve 2×2.
Answer (x,y,z):
D
Special Cases — I.M.S. & Inconsistent
Q10
Trap!

Is this system Inconsistent, I.M.S., or has a unique solution?

\(3x - 3y + 6z = 6\)
\(x + 2y - z = 5\)
\(5x - 8y + 13z = 7\)
⚡ Fast Method Eliminate x: ①−3×②, ③−5×② → two new equations.
If they reduce to 0=0 → I.M.S.   If 0=k → Inconsistent.
Type?
Q11
Trap!

Is this system Inconsistent, I.M.S., or unique?

\(3x - 2y + 4z = 1\)
\(x + y - 2z = 3\)
\(2x - 3y + 6z = 8\)
⚡ Fast Method Same approach. After elimination if you reach 0 = constant (≠0) it's inconsistent.
Type?
E
Solve — Harder Systems
Q12
Medium

Solve the system.

\(3x - y + z = 1\)
\(x + y - 2z = 3\)
\(2x - 3y + 6z = 8\)
⚡ Fast Method Add ①+②: eliminate y → \(4x-z=4\).
Multiply ②×3 add to ③: eliminate y → \(5x=17\).
Solve x, back-sub for z, then y.
Answer (x,y,z):
Q13
Medium

Solve the system.

\(x + 2y + z = 1\)
\(x - 2y + 3z = -3\)
\(2x + y + z = -1\)
⚡ Fast Method ①−②: \(4y-2z=4\). ①+②: \(2x+4z=-2\Rightarrow x+2z=-1\). Combine with ③ to find z first.
Answer (x,y,z):
Q14
Hard

Solve the system.

\(4x + y - 3z = 11\)
\(2x - 3y + 2z = 9\)
\(x + y + z = -3\)
⚡ Fast Method Multiply eq③×4, subtract from ①: removes x → eq in y,z.
Multiply eq③×2, subtract from ②: another y,z eq. Then solve 2×2.
Answer (x,y,z):
Q15
Hard

Solve the system.

\(x + 2y + z = 1\)
\(x - 2y + 3z = -3\)
\(2x + y + z = -1\)
⚡ Fast Method Subtract ②from①: \(4y-2z=4\Rightarrow 2y-z=2\).
Subtract ③from①: \(y=2\). Then z=2, x from eq①.
Answer (x,y,z):
F
Concept & Vocab
Q16
Concept

When you eliminate variables and end up with 0 = 0, what does it mean?

⚡ Key Rule 0 = 0 means the equations are dependent (same plane intersecting). The system has infinitely many solutions.
Answer:
Q17
Concept

When elimination gives 0 = 5, what is the solution set?

⚡ Key Rule 0 = k (k ≠ 0) is a contradiction. The planes never meet. No solution exists.
Answer:
Q18
Medium

Solve and write the ordered triple.

\(3x - y + z = 1\)
\(x + y - 2z = 3\)
\(x + 3y - z = 5\)
⚡ Fast Method ①+②: \(4x-z=4\). ②+③ eliminate y: \(2x-3z=8\).
Solve 2×2 for x and z, then back-sub for y.
Answer (x,y,z):
Q19
Hard

Solve the system.

\(2x + y - z = 4\)
\(x - y + 2z = -1\)
\(3x + 2y - z = 6\)
⚡ Fast Method ①+②: \(3x+z=3\). ①×2−③: \(x-z=2\).
Add both new eqs: \(4x=5\)? Solve carefully for x, z, then y.
Answer (x,y,z):
Q20
Challenge

Solve the system — watch for a special case!

\(x + 2y - z = 5\)
\(2x - y + 3z = 0\)
\(3x + y + 2z = 5\)
⚡ Fast Method ①×2−②: eliminate x → \(5y-5z=10\Rightarrow y-z=2\).
①×3−③: another eq in y,z → compare. Then solve back.
Answer (x,y,z):

You finished! 🎉

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out of 20 correct