Lesson 9 of 42 · Algebra

Systems of equations with two unknowns

Solve two-unknown systems by substitution or elimination, tell when there is one, no or infinitely many solutions, and intersect a line with a parabola.

About 22 minutes

In this lesson

  • Solve a linear system in two unknowns by substitution or by elimination, choosing the quicker method.
  • Decide from the coefficients whether a linear system has one solution, none or infinitely many.
  • Solve a system of a line and a parabola, and find what a question asks for without solving fully when a combination is enough.

The idea

What a solution is

A solution of a system is a pair (x,y)(x, y) that satisfies every equation at once. Each linear equation ax+by=cax + by = c is a straight line in the plane, so solving two of them means finding where the two lines meet.

Substitution

Isolate one unknown in the equation where it is easiest (a coefficient of 11 or −1-1), and substitute into the other:

{x−y=12x+3y=7⇒x=1+y⇒2(1+y)+3y=7.\begin{cases} x - y = 1 \\ 2x + 3y = 7 \end{cases} \quad\Rightarrow\quad x = 1 + y \quad\Rightarrow\quad 2(1 + y) + 3y = 7.

So 5y=55y = 5, y=1y = 1 and x=2x = 2. Substitution is the natural choice when one equation is not linear, for example a parabola and a line.

Elimination

Multiply the equations by numbers that make the coefficients of one unknown equal or opposite, then add or subtract the equations to remove it. For

{3x+2y=165x−2y=8\begin{cases} 3x + 2y = 16 \\ 5x - 2y = 8 \end{cases}

adding removes yy at once: 8x=248x = 24. Elimination is usually faster when both equations are linear and no coefficient is 11.

Combinations: answer the question asked

A question may ask for an expression such as x+yx + y or 3x+2y3x + 2y, not for xx and yy. Before solving, see whether adding or subtracting the equations produces that expression directly. In the system above, the sum of the two equations is 8x=248x = 24; in the first system, adding x−y=1x - y = 1 to 2x+3y=72x + 3y = 7 gives 3x+2y=83x + 2y = 8 immediately.

One, none or infinitely many

Two lines in the plane meet once, never (parallel) or everywhere (the same line). For

{a1x+b1y=c1a2x+b2y=c2\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}

the test is the number a1b2−a2b1a_1 b_2 - a_2 b_1:

Condition Lines Solutions
a1b2−a2b1≠0a_1 b_2 - a_2 b_1 \ne 0 cross exactly one
a1b2−a2b1=0a_1 b_2 - a_2 b_1 = 0, equations inconsistent parallel, distinct none
a1b2−a2b1=0a_1 b_2 - a_2 b_1 = 0, one equation a multiple of the other the same line infinitely many

When a1b2−a2b1=0a_1 b_2 - a_2 b_1 = 0, the left sides are proportional. Compare the right sides with the same factor: if they match, the equations say the same thing; if not, they contradict each other. For example, x+2y=4x + 2y = 4 and 3x+6y=123x + 6y = 12 are the same line, while x+2y=4x + 2y = 4 and 3x+6y=53x + 6y = 5 are parallel.

A line and a parabola

To intersect y=x2+px+qy = x^2 + px + q with y=mx+ny = mx + n, substitute the line into the parabola. You get a quadratic in xx, and its discriminant tells you how the line sits:

  • Δ>0\Delta > 0: two intersection points (a secant);
  • Δ=0\Delta = 0: one point, the line is tangent;
  • Δ<0\Delta < 0: no common point.

Each root xx gives a point; find its yy from the line, which is quicker, and check it in the parabola.

Word problems

Name the two unknowns explicitly, with units ("aa = number of adult tickets"), write one equation per independent fact, solve, then check the answer against the story: counts must be whole numbers, lengths positive. The check catches most setting-up errors.

Worked examples

Example 1: Elimination in one step

Solve 3x+2y=163x + 2y = 16 and 5x−2y=85x - 2y = 8.

  1. The yy coefficients are opposite, so add the equations: 8x=248x = 24, x=3x = 3.

  2. Substitute into the first: 9+2y=169 + 2y = 16, so y=3.5y = 3.5.

  3. Check in the second: 15−7=815 - 7 = 8.

Answer

(x,y)=(3,3.5)(x, y) = (3, 3.5). Subtracting the equations instead of adding them gives −2x+4y=8-2x + 4y = 8, which still has both unknowns and solves nothing.

Example 2: Counting solutions with a parameter

For which values of kk does the system kx+2y=4kx + 2y = 4, 3x+6y=53x + 6y = 5 have exactly one solution? What happens for the other values?

  1. Compute a1b2−a2b1=k⋅6−3⋅2=6k−6a_1 b_2 - a_2 b_1 = k \cdot 6 - 3 \cdot 2 = 6k - 6.

  2. Exactly one solution when 6k−6≠06k - 6 \ne 0, that is k≠1k \ne 1.

  3. At k=1k = 1: the first equation is x+2y=4x + 2y = 4; dividing the second by 3 gives x+2y=53x + 2y = \frac{5}{3}. The same left side cannot equal both 4 and 53\frac{5}{3}, so there is no solution.

Answer

One solution for every k≠1k \ne 1; no solution at k=1k = 1, where the lines are parallel and distinct. Concluding "infinitely many" at k=1k = 1 skips the comparison of the right sides.

Example 3: Where a line meets a parabola

Find the intersection points of y=x2−2xy = x^2 - 2x and y=x+4y = x + 4.

  1. Set the two expressions for yy equal: x2−2x=x+4x^2 - 2x = x + 4, so x2−3x−4=0x^2 - 3x - 4 = 0.

  2. Factor: (x−4)(x+1)=0(x - 4)(x + 1) = 0, so x=4x = 4 or x=−1x = -1.

  3. Find yy from the line: x=4x = 4 gives y=8y = 8; x=−1x = -1 gives y=3y = 3.

  4. Check in the parabola: 16−8=816 - 8 = 8 and 1+2=31 + 2 = 3.

Answer

The points are (4,8)(4, 8) and (−1,3)(-1, 3). Stopping at x=4x = 4 and x=−1x = -1 gives only half of each answer; a point needs both coordinates.

Example 4: A word problem

A theatre sold 40 tickets for €395. Adult tickets cost €12 and child tickets €7. How many of each were sold?

  1. Let aa be the number of adult tickets and cc the number of child tickets: a+c=40a + c = 40 and 12a+7c=39512a + 7c = 395.

  2. Substitute c=40−ac = 40 - a: 12a+280−7a=39512a + 280 - 7a = 395, so 5a=1155a = 115 and a=23a = 23.

  3. Then c=17c = 17. Check the money: 12⋅23+7⋅17=276+119=39512 \cdot 23 + 7 \cdot 17 = 276 + 119 = 395.

Answer

23 adult and 17 child tickets. If a setup gives a fraction of a ticket, the equations are wrong, not the arithmetic.

Common traps

  • Substitution sign slips

    From x−y=1x - y = 1 the substitution is x=1+yx = 1 + y, not x=1−yx = 1 - y. After solving, put the pair back into both original equations; one check is not enough.

  • Assuming a proportional system has no solution

    When the left sides are proportional, compare the right sides with the same factor. Matching right sides mean the same line and infinitely many solutions; only mismatched ones mean none.

  • Answering with x when the question asks for something else

    Read what is asked: yy, x+yx + y, the product xyxy, or the number of solutions. Often a combination of the equations gives it directly, and finding xx alone answers nothing.

  • Taking y from the wrong curve

    For a line and a parabola both curves give the same yy at a true intersection point, so either works, but a slip in squaring a negative xx is common. Use the line for speed and the parabola as a check.

Check yourself

Answer each question before you open its answer.

Question 1: Asked for a combination

Algebra: systems

If 2x+3y=72x + 3y = 7 and x−y=1x - y = 1, what is the value of 3x+2y3x + 2y?

  1. A.6
  2. B.7
  3. C.8
  4. D.9
Show the answer

Answer: C. 8

Working

Add the two equations: (2x+3y)+(x−y)=7+1(2x + 3y) + (x - y) = 7 + 1, so 3x+2y=83x + 2y = 8 directly. Check by solving: x=1+yx = 1 + y gives 2+2y+3y=72 + 2y + 3y = 7, so y=1y = 1, x=2x = 2, and 3(2)+2(1)=83(2) + 2(1) = 8.

Why the other options are wrong

  • A. This subtracts the equations, which gives x+4y=6x + 4y = 6: a different combination from the one asked.
  • B. This swaps the solution, using (x,y)=(1,2)(x, y) = (1, 2); that pair does not satisfy x−y=1x - y = 1.
  • D. This is 3(x+y)=3×33(x + y) = 3 \times 3, using 3 as the coefficient of yy as well as of xx.

What carries over

Look for a combination of the equations that gives the requested expression before solving for each unknown.

Question 2: A system with a parameter

Algebra: systems

Consider the system 2x+ay=32x + ay = 3 and 4x+6y=64x + 6y = 6. Which statement is true?

  1. A.At a=3a = 3 the system has no solution.
  2. B.At a=3a = 3 the system has infinitely many solutions.
  3. C.The system has exactly one solution for every value of aa.
  4. D.At a=12a = 12 the system has infinitely many solutions.
Show the answer

Answer: B. At a=3a = 3 the system has infinitely many solutions.

Working

a1b2−a2b1=2⋅6−4a=12−4aa_1 b_2 - a_2 b_1 = 2 \cdot 6 - 4a = 12 - 4a, which is zero only at a=3a = 3; for every other aa there is exactly one solution. At a=3a = 3 the first equation is 2x+3y=32x + 3y = 3, and twice it is 4x+6y=64x + 6y = 6, the second equation exactly. The two equations describe the same line, so there are infinitely many solutions.

Why the other options are wrong

  • A. At a=3a = 3 the second equation is exactly twice the first (4x+6y=64x + 6y = 6), so the lines coincide; there are infinitely many solutions, not none.
  • C. At a=3a = 3 the expression a1b2−a2b1=12−4aa_1 b_2 - a_2 b_1 = 12 - 4a is zero, so the solution is not unique there.
  • D. This comes from a cross-multiplication slip, setting a=2⋅6a = 2 \cdot 6. The left sides are proportional when 24=a6\frac{2}{4} = \frac{a}{6}, which gives a=3a = 3.

What carries over

When the left sides are proportional, compare the right sides to decide between none and infinitely many.

Question 3: Line meets parabola

Algebra: systems

The line y=2x+3y = 2x + 3 meets the parabola y=x2y = x^2 at two points. What is the sum of the yy-coordinates of these points?

  1. A.−3-3
  2. B.22
  3. C.88
  4. D.1010
Show the answer

Answer: D. 1010

Working

Set x2=2x+3x^2 = 2x + 3: x2−2x−3=0x^2 - 2x - 3 = 0, so (x−3)(x+1)=0(x - 3)(x + 1) = 0 and x=3x = 3 or x=−1x = -1. The points are (3,9)(3, 9) and (−1,1)(-1, 1); check with the line: 2(3)+3=92(3) + 3 = 9 and 2(−1)+3=12(-1) + 3 = 1. The sum of the yy-coordinates is 9+1=109 + 1 = 10.

Why the other options are wrong

  • A. This is the product of the xx-coordinates, 3×(−1)3 \times (-1), not a sum of yy-coordinates.
  • B. This is the sum of the xx-coordinates, 3+(−1)3 + (-1).
  • C. This takes y=−1y = -1 at x=−1x = -1, forgetting that (−1)2=1(-1)^2 = 1; the line agrees: 2(−1)+3=12(-1) + 3 = 1.

What carries over

Substitute to get a quadratic in x, then find each y and check it on both curves.

Key points

  • Use substitution when one equation is easy to rearrange or not linear; elimination when both are linear.

  • a1b2−a2b1≠0a_1 b_2 - a_2 b_1 \ne 0 means exactly one solution; if it is zero, compare the right sides: same line or parallel lines.

  • A line and a parabola meet in two, one or no points, according to the discriminant of the resulting quadratic.

  • Read what is asked: a combination of the equations often gives it without solving.

All questions here are original and written for UniExamer; none comes from the test maker.

Written by the UniExamer team with AI assistance. Every worked example and check question was re-solved by an independent checker before publication. No subject expert has reviewed this lesson yet. If you find a mistake, write to support@uniexamer.com.

Esercitati su questo argomento

Domande nuove su questa abilità, con spiegazioni svolte. L’esercitazione richiede un Single Exam Pass o Plus.