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Systems in Two Variables: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Two shifted quantities

Find the real pair $(u,v)$ satisfying $2(u-1)=v+3$ and $u+v=7$ .
1. Hint 1
  Rewrite the first condition so that its variable terms are together.
2. Hint 2
  Add the two conditions to eliminate one variable.
Answer

$(u,v)=(4,3)$ .

Full solution

The first condition gives $2u-v=5$ .
Add it to the sum condition.
$\begin{aligned} 3u&=12 \\ u&=4 \\ v&=7-4 \end{aligned}$

Thus $v=3$ .
Check: $2(4-1)=6$ and $3+3=6$ , and $4+3=7$ .

Answer

$(u,v)=(4,3)$ .

Key idea

A pair solves a system when it meets every condition at once.
Problem 1: Mark complete
Problem 2 Two recorded totals

Two instruments report $4a+6b=18$ and $2a+3b=10$ . Determine whether any real settings $(a,b)$ fit both reports.
1. Hint 1
  Compare the measured expressions before looking for individual settings.
2. Hint 2
  Double the report with the smaller coefficients.
Answer

No settings fit; the system is inconsistent.

Full solution

Doubling the second report gives $4a+6b=20$ .
Subtract the first report.
$0=2$
No real settings can make this true.

Answer

No settings fit; the system is inconsistent.

Key idea

Conflicting totals for the same expression make a system inconsistent.
Problem 2: Mark complete
Problem 3 A family of pairs

Write all solutions of $y=5-\frac{x}{2}$ and $x=10-2y$ as ordered pairs using a real parameter $t$ .
1. Hint 1
  An equivalent rearrangement may reveal that there is just one restriction.
2. Hint 2
  Let $y=t$ and find the matching first coordinate.
Answer

$(x,y)=(10-2t,t)$ for every real $t$ (equivalently $(t,5-\tfrac{t}{2})$ , any parametrization of the same line $x+2y=10$ ).

Full solution

Multiplying the first equation by $2$ and rearranging gives the second.
Set $y=t$ .
$x=10-2t$
Substitution returns $5-(10-2t)/2=t$ , so every real $t$ works and every solution is included.

Answer

$(x,y)=(10-2t,t)$ for every real $t$ (equivalently $(t,5-\tfrac{t}{2})$ , any parametrization of the same line $x+2y=10$ ).

Key idea

One surviving linear restriction leaves one freely chosen coordinate.
Problem 3: Mark complete
Problem 4 Coefficients on a card

A card lists matching coefficient ratios $a_1/a_2=3/5$ and $b_1/b_2=3/5$ , with all six coefficients and constants nonzero. The constant ratio is $c_1/c_2=2/5$ . Classify the system $a_ix+b_iy=c_i$ and explain.
1. Hint 1
  Compare the scaling of the whole equations, not just the variable terms.
2. Hint 2
  Scale the first equation by $5$ and the second by $3$ .
Answer

Inconsistent; no solution.

Full solution

The scaled variable terms agree because $5a_1=3a_2$ and $5b_1=3b_2$ .
But $c_1=2c_2/5$ , so subtracting the scaled equations gives
$0=-c_2$
This is false since $c_2\ne0$ .
The lines are parallel and distinct.

Answer

Inconsistent; no solution.

Key idea

Equal coefficient ratios require the same constant ratio for consistency.
Problem 4: Mark complete
Problem 5 Calibration readings

A scale displays $d=kw+b$ , where $w$ is the actual mass in grams and $d$ is the displayed number. A 2 gram mass displays 7, and a 5 gram mass displays 13. Find $k$ and $b$ , and classify the resulting system.
1. Hint 1
  Each measurement gives one condition on the same two settings.
2. Hint 2
  Subtract the conditions to remove the offset.
Answer

$k=2$ , $b=3$ ; consistent and independent.

Full solution

$\begin{aligned} 2k+b&=7 \\ 5k+b&=13 \end{aligned}$

Subtraction gives $3k=6$ , so $k=2$ .
Then $b=7-4=3$ .
Both readings check: $2(2)+3=7$ and $2(5)+3=13$ .
There is one pair of settings, so the system is consistent and independent.

Answer

$k=2$ , $b=3$ ; consistent and independent.

Key idea

Two distinct inputs can determine the slope and offset of a linear rule.
Problem 5: Mark complete
Problem 6 A movable relation

For real $k$ , consider $y=kx+2$ and $y=(k+1)x+2$ . Determine the solution and classification for every value of $k$ .
1. Hint 1
  Compare the equations without dividing by $k$ .
2. Hint 2
  Subtract their right sides at a common point.
Answer

$(x,y)=(0,2)$ for every real $k$ ; consistent and independent.

Full solution

At a common point, the two right sides agree.
$\begin{aligned} kx+2&=(k+1)x+2 \\ kx+2&=kx+x+2 \\ 0&=x \end{aligned}$
Expanding $(k+1)x$ as $kx+x$ and canceling $kx+2$ from both sides leaves $0=x$ .

Substitution gives $y=2$ .
The slopes differ by $1$ for every $k$ , so this is a single intersection, including $k=0$ .

Answer

$(x,y)=(0,2)$ for every real $k$ ; consistent and independent.

Key idea

A parameter may change both lines while leaving their unique intersection fixed.
Problem 6: Mark complete
Problem 7 Shared marks

Two genuine linear equations in $x,y$ are both satisfied by $(1,4)$ and $(3,0)$ . Find the entire solution set and classify the system.
1. Hint 1
  Two distinct points determine a line.
2. Hint 2
  Use the change in height over the change in horizontal position.
Answer

$(x,y)=(t,6-2t)$ for all real $t$ ; consistent and dependent.

Full solution

The slope through the given points is
$(0-4)/(3-1)=-2.$
The point $(1,4)$ fixes the intercept.
$y=6-2x$
Each equation describes the unique line through those points, so both describe this line.
Setting $x=t$ lists every shared point.

Answer

$(x,y)=(t,6-2t)$ for all real $t$ ; consistent and dependent.

Key idea

Two shared points force two genuine lines to coincide.
Problem 7: Mark complete
Problem 8 A calculation record

A student keeps $x+2y=9$ and replaces $3x+7y=30$ by its difference from three times the kept equation. They get $y=3$ and announce infinitely many solutions because one variable disappeared. Is the conclusion correct? Explain and give the actual solution.
1. Hint 1
  Removing a variable from one equation is different from losing a constraint.
2. Hint 2
  Use the remaining one-variable equation in the kept equation.
Answer

No; the unique solution is $(3,3)$ .

Full solution

The replacement is correct:
$y=30-3(9)$
Hence $y=3$ .
Substituting in the kept equation gives $x+6=9$ , so $x=3$ .
Both original equations hold: $3+6=9$ and $9+21=30$ .
A genuine equation for $y$ still constrains it.

Answer

No; the unique solution is $(3,3)$ .

Key idea

Elimination that leaves a nonzero variable coefficient determines a value rather than a free variable.
Problem 8: Mark complete
Problem 9 A segment of solutions

Two genuine linear equations have every point of a nontrivial line segment as solutions. A student says the solution set must include the entire line containing that segment. Is this correct? Justify.
1. Hint 1
  A nontrivial segment supplies two distinct shared points.
2. Hint 2
  Compare the line determined by those two points with each equation.
Answer

Yes; the full containing line is the solution set.

Full solution

Choose distinct points $P$ and $Q$ of the segment.
Each equation describes a line through both, and there is just one line through two distinct points.
Thus both equations describe that same entire line.
Every point on it satisfies both equations, and points off it satisfy neither.

Answer

Yes; the full containing line is the solution set.

Key idea

Two linear equations that share a segment share its entire containing line.
Problem 9: Mark complete
Problem 10 A vertical family

For the system $3x=12$ and $9x=36$ in the variables $x$ and $y$ , someone lists $(4,t)$ only for integer $t$ . Does that give every real solution? Explain and correct the family if needed.
1. Hint 1
  Neither equation contains $y$ .
2. Hint 2
  Check a noninteger value of the second coordinate.
Answer

No; all solutions are $(4,t)$ for every real $t$ .

Full solution

Both equations reduce to
$x=4$
They put no restriction on $y$ .
For example, $(4,1/2)$ solves both but is absent from the proposed integer list.
Allowing every real $t$ supplies precisely the whole vertical line.

Answer

No; all solutions are $(4,t)$ for every real $t$ .

Key idea

A free coordinate ranges over the stated domain, not just convenient sample values.
Problem 10: Mark complete

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