Systems of Equations

Study Sheet

Systems of Equations

Everything you need: graphing, substitution, elimination, special cases, inequalities, and word problems.

What Is a System of Equations?

Concept
The Big Idea

A system of equations is two (or more) equations that we want to be true at the same time. Each linear equation is a line on the coordinate plane.

A solution to the system is an ordered pair (x,y)(x,y) that makes every equation true. Graphically, it is the point where the lines cross.

We usually write a system with a curly brace:

{x+y=6x−y=2\begin{cases} x + y = 6 \\ x - y = 2 \end{cases}

This means: find the one pair (x,y)(x,y) that fits both equations.

Example
Checking a proposed solution

Is (4,2)(4,2) a solution of the system above?

  • First equation: x+y=4+2=6x+y = 4+2 = 6. ✓
  • Second equation: x−y=4−2=2x-y = 4-2 = 2. ✓

Both are true, so (4,2)(4,2) is the solution. If even one equation failed, it would not be a solution.

Tip

Remember: A solution must satisfy all equations. Always plug your answer back into both equations to check.

Method 1: Solving by Graphing

Concept
How Graphing Works

Graph both lines on the same axes. The intersection point (x,y)(x,y) is the solution, because that point lies on both lines at once.

Steps: (1) Put each equation in slope-intercept form y=mx+by = mx + b if helpful. (2) Plot each line using its yy-intercept and slope (or a small table of points). (3) Read off the point where they cross. (4) Check by substituting.

Example
Graphing example

Solve {y=x+1y=−2x+4\begin{cases} y = x + 1 \\ y = -2x + 4 \end{cases}

Line 1: y=x+1y = x+1 passes through (0,1)(0,1) and (1,2)(1,2).

Line 2: y=−2x+4y = -2x+4 passes through (0,4)(0,4) and (1,2)(1,2).

Both tables contain the point (1,2)(1,2), so the lines cross there.

Check: y=x+1=1+1=2y = x+1 = 1+1 = 2 ✓  and   y=−2(1)+4=2y = -2(1)+4 = 2 ✓

Solution: (1,2)(1,2).

Tip

Tip: Graphing shows the answer at a glance, but it is only exact when the intersection lands on a clean grid point. For messy or fractional answers, use substitution or elimination instead.

Method 2: Solving by Substitution

Concept
How Substitution Works

Solve one equation for one variable, then substitute that expression into the other equation. Now you have one equation with one variable, which you can solve. Finally, back-substitute to find the other variable.

Best when: one variable is already isolated (like y=…y = \dots) or is easy to isolate.

Example
Substitution, fully stepped out

Solve {y=2x+13x+y=11\begin{cases} y = 2x + 1 \\ 3x + y = 11 \end{cases}

Step 1. The first equation already gives y=2x+1y = 2x+1.

Step 2. Substitute 2x+12x+1 in place of yy in the second equation:

3x+(2x+1)=113x + (2x + 1) = 11

Step 3. Combine and solve for xx:

5x+1=11  ⇒  5x=10  ⇒  x=25x + 1 = 11 \;\Rightarrow\; 5x = 10 \;\Rightarrow\; x = 2

Step 4. Back-substitute x=2x=2 into y=2x+1y = 2x+1:

y=2(2)+1=5y = 2(2) + 1 = 5

Check: 3(2)+5=113(2) + 5 = 11 ✓

Solution: (2,5)(2,5).

Example
Substitution needing rearrangement

Solve {x−3y=−12x+y=5\begin{cases} x - 3y = -1 \\ 2x + y = 5 \end{cases}

Step 1. Isolate xx in the first equation: x=3y−1x = 3y - 1.

Step 2. Substitute into the second: 2(3y−1)+y=52(3y-1) + y = 5.

Step 3. Solve: 6y−2+y=5⇒7y=7⇒y=16y - 2 + y = 5 \Rightarrow 7y = 7 \Rightarrow y = 1.

Step 4. Then x=3(1)−1=2x = 3(1) - 1 = 2.

Check: 2(2)+1=52(2) + 1 = 5 ✓   Solution: (2,1)(2,1).

Tip

Watch out: Use parentheses when you substitute, especially with a minus sign in front. Distribute carefully.

Method 3: Solving by Elimination (Linear Combination)

Concept
How Elimination Works

Add or subtract the two equations so that one variable cancels. To make a variable cancel, its coefficients must be opposites (like +2y+2y and −2y-2y). If they are not, multiply one or both equations by a number first.

Best when: the equations are in the form ax+by=cax + by = c and no variable is isolated.

Example
Elimination when coefficients already match

Solve {x+y=10x−y=4\begin{cases} x + y = 10 \\ x - y = 4 \end{cases}

The +y+y and −y-y are already opposites. Add the equations:

(x+y)+(x−y)=10+4  ⇒  2x=14  ⇒  x=7(x + y) + (x - y) = 10 + 4 \;\Rightarrow\; 2x = 14 \;\Rightarrow\; x = 7

Back-substitute into x+y=10x + y = 10:   7+y=10⇒y=3\;7 + y = 10 \Rightarrow y = 3.

Check: 7−3=47 - 3 = 4 ✓   Solution: (7,3)(7,3).

Example
Elimination needing multiplication (one equation)

Solve {2x+3y=74x−y=7\begin{cases} 2x + 3y = 7 \\ 4x - y = 7 \end{cases}

Step 1. Choose a variable to eliminate. Multiply the first equation by 22 so the xx-terms match:

2(2x+3y)=2(7)  ⇒  4x+6y=142(2x + 3y) = 2(7) \;\Rightarrow\; 4x + 6y = 14

Step 2. Subtract the second equation from this new one:

(4x+6y)−(4x−y)=14−7  ⇒  7y=7  ⇒  y=1(4x + 6y) - (4x - y) = 14 - 7 \;\Rightarrow\; 7y = 7 \;\Rightarrow\; y = 1

Step 3. Back-substitute into 2x+3y=72x + 3y = 7:   2x+3=7⇒x=2\;2x + 3 = 7 \Rightarrow x = 2.

Check: 4(2)−1=74(2) - 1 = 7 ✓   Solution: (2,1)(2,1).

Example
Elimination needing multiplication (both equations)

Solve {3x+4y=102x+3y=7\begin{cases} 3x + 4y = 10 \\ 2x + 3y = 7 \end{cases}

To cancel xx, make both xx-coefficients 66. Multiply the first by 22 and the second by 33:

6x+8y=206x+9y=216x + 8y = 20 \qquad 6x + 9y = 21

Subtract: (6x+8y)−(6x+9y)=20−21⇒−y=−1⇒y=1(6x+8y) - (6x+9y) = 20 - 21 \Rightarrow -y = -1 \Rightarrow y = 1.

Back-substitute into 3x+4y=103x + 4y = 10:   3x+4=10⇒x=2\;3x + 4 = 10 \Rightarrow x = 2.

Check: 2(2)+3(1)=72(2) + 3(1) = 7 ✓   Solution: (2,1)(2,1).

Tip

Tips for elimination: Line up like variables in columns before you start. When you subtract, change the sign of every term in the equation you subtract. Double-check signs, since that is the most common mistake.

How Many Solutions? Three Possibilities

Concept
One, None, or Infinitely Many

Two lines can meet in exactly three ways:

  • One solution: the lines cross once (different slopes). This is the usual case.
  • No solution: the lines are parallel (same slope, different yy-intercept). They never meet.
  • Infinitely many solutions: the two equations are really the same line (same slope and same intercept). Every point on the line works.
Example
Spotting the special cases

No solution: {y=2x+1y=2x−3\begin{cases} y = 2x + 1 \\ y = 2x - 3 \end{cases} Same slope 22, different intercepts ⇒\Rightarrow parallel ⇒\Rightarrow no solution.

Infinitely many: {2x+y=44x+2y=8\begin{cases} 2x + y = 4 \\ 4x + 2y = 8 \end{cases} The second equation is just the first times 22, so they are the same line ⇒\Rightarrow infinitely many solutions.

Tip

Algebra clue: If your work leads to a false statement like 0=60 = 6, there is no solution. If it leads to an always-true statement like 0=00 = 0, there are infinitely many solutions.

Choosing the Best Method

Concept
Which Method Should I Use?
  • Graphing --- great for a quick picture or when you want to see the answer; best with clean integer intersections.
  • Substitution --- best when a variable is already isolated or has a coefficient of 11 (easy to solve for).
  • Elimination --- best when both equations are in ax+by=cax + by = c form, especially if coefficients line up or match easily.

There is no single “right” method; pick whichever needs the least work for the problem in front of you.

Systems of Linear Inequalities

Concept
Shading the Solution Region

A system of inequalities uses <, >, ≤, ≥<,\,>,\,\le,\,\ge instead of ==. Instead of a single point, the solution is a whole region of the plane.

Steps: (1) Graph each boundary line. Use a solid line for ≤\le or ≥\ge and a dashed line for << or >>. (2) Shade the correct side of each line. (3) The solution is the overlap where the shadings cross. Any point in that overlap satisfies both inequalities.

Example
Testing points in an inequality system

Consider {y≥x−1y<−x+3\begin{cases} y \ge x - 1 \\ y < -x + 3 \end{cases}

Use a test point to decide which side to shade and to check membership.

Test (0,0)(0,0):   0≥0−1\;0 \ge 0 - 1 (true) and 0<0+30 < 0 + 3 (true). Both hold, so (0,0)(0,0) is in the solution region.

Test (3,3)(3,3):   3≥3−1=2\;3 \ge 3 - 1 = 2 (true) but 3<−3+3=03 < -3 + 3 = 0 (false). One fails, so (3,3)(3,3) is not in the region.

The solution is the overlapping band above the line y=x−1y = x-1 and below the line y=−x+3y = -x+3.

Tip

Tip: (0,0)(0,0) is the easiest test point to plug in, as long as no boundary line passes through the origin. A point is a solution only if it satisfies every inequality.

Applications: Word Problems

Concept
Turning Words into a System

Many word problems hide a 2×22\times2 system. Steps: (1) Define two variables clearly. (2) Write one equation for each fact in the problem. (3) Solve with any method. (4) Answer in a full sentence and check that it makes sense.

Example
Cost / number problem

A theater sells $5 adult tickets and $3 child tickets. One night they sold 1010 tickets for a total of $38. How many of each?

Define: a=a = adult tickets, c=c = child tickets.

System: {a+c=105a+3c=38\begin{cases} a + c = 10 \\ 5a + 3c = 38 \end{cases}

Solve by substitution: a=10−ca = 10 - c, so 5(10−c)+3c=38⇒50−2c=38⇒c=65(10-c) + 3c = 38 \Rightarrow 50 - 2c = 38 \Rightarrow c = 6, then a=4a = 4.

Check: 5(4)+3(6)=20+18=385(4) + 3(6) = 20 + 18 = 38 ✓   Answer: 44 adult and 66 child tickets.

Example
Mixture problem

A chemist mixes a 20%20\% acid solution with a 50%50\% acid solution to make 3030 mL of a 40%40\% acid solution. How much of each?

Define: x=x = mL of 20%20\% solution, y=y = mL of 50%50\% solution.

System: {x+y=300.20x+0.50y=0.40(30)=12\begin{cases} x + y = 30 \\ 0.20x + 0.50y = 0.40(30) = 12 \end{cases}

Solve: x=30−yx = 30 - y, so 0.20(30−y)+0.50y=12⇒6+0.30y=12⇒y=200.20(30-y) + 0.50y = 12 \Rightarrow 6 + 0.30y = 12 \Rightarrow y = 20, then x=10x = 10.

Check: 0.20(10)+0.50(20)=2+10=120.20(10) + 0.50(20) = 2 + 10 = 12 ✓   Answer: 1010 mL of 20%20\% and 2020 mL of 50%50\%.

Tip

Tip: Distance--rate--time and current problems often become systems too. If a boat's speed is bb and the current is cc, then downstream speed is b+cb+c and upstream speed is b−cb-c. Two trips give you two equations.

Going Deeper: Advanced Systems

Concept
Systems in Three Variables

A three-variable system has three equations in xx, yy, and zz. A solution is an ordered triple (x,y,z)(x,y,z) that satisfies all three. Geometrically each equation is a plane in space, and a single solution is the point where all three planes meet.

Elimination strategy: pick one variable to remove. Combine equations in two different pairs to eliminate that same variable, leaving a 2×22\times2 system in the other two variables. Solve that smaller system, then back-substitute to recover the third variable.

Example
Three-variable elimination, fully stepped out

Solve

{x+y+z=6x−y+z=22x+y−z=1\begin{cases} x + y + z = 6 \\ x - y + z = 2 \\ 2x + y - z = 1 \end{cases}

Step 1. Eliminate zz from two pairs. Add equation (1) and equation (3):

(x+y+z)+(2x+y−z)=6+1  ⇒  3x+2y=7(x+y+z) + (2x+y-z) = 6 + 1 \;\Rightarrow\; 3x + 2y = 7

Add equation (2) and equation (3):

(x−y+z)+(2x+y−z)=2+1  ⇒  3x=3  ⇒  x=1(x-y+z) + (2x+y-z) = 2 + 1 \;\Rightarrow\; 3x = 3 \;\Rightarrow\; x = 1

Step 2. Back-substitute x=1x=1 into 3x+2y=73x + 2y = 7:

3(1)+2y=7  ⇒  2y=4  ⇒  y=23(1) + 2y = 7 \;\Rightarrow\; 2y = 4 \;\Rightarrow\; y = 2

Step 3. Find zz from equation (1):   1+2+z=6⇒z=3\;1 + 2 + z = 6 \Rightarrow z = 3.

Check (all three): 1+2+3=61+2+3=6 ✓,   1−2+3=2\;1-2+3=2 ✓,   2(1)+2−3=1\;2(1)+2-3=1 ✓

Solution: (x,y,z)=(1,2,3)(x,y,z) = (1,2,3).

Concept
Systems With a Parameter

Sometimes a coefficient is a letter (a parameter) and we ask: for which values does the system have one, none, or infinitely many solutions? Compare the ratios of the coefficients. 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}
  • One solution when a1a2≠b1b2\dfrac{a_1}{a_2} \ne \dfrac{b_1}{b_2} (different slopes).
  • Infinitely many when a1a2=b1b2=c1c2\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2} (same line).
  • No solution when a1a2=b1b2≠c1c2\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \ne \dfrac{c_1}{c_2} (parallel lines).
Example
Choosing the parameter

For which value of kk does {x+2y=32x+ky=m\begin{cases} x + 2y = 3 \\ 2x + ky = m \end{cases} fail to have a unique solution?

The xx-ratio is 12\dfrac{1}{2}. The yy-ratio is 2k\dfrac{2}{k}. These match when 2k=12\dfrac{2}{k} = \dfrac{1}{2}, i.e. k=4k = 4. So:

  • If k≠4k \ne 4: exactly one solution for any mm.
  • If k=4k = 4 and m=6m = 6: the second equation is just 2×2\times the first, so infinitely many solutions.
  • If k=4k = 4 and m≠6m \ne 6: parallel lines, so no solution.
Concept
Symmetric Systems: the x+yx+y and xyxy Trick

A system is symmetric if swapping xx and yy leaves it unchanged (for example, using x+yx+y, xyxy, or x2+y2x^2+y^2). Let s=x+ys = x + y (the sum) and p=xyp = xy (the product). A key identity converts squares into these:

x2+y2=(x+y)2−2xy=s2−2px^2 + y^2 = (x+y)^2 - 2xy = s^2 - 2p

Once you know ss and pp, the two numbers xx and yy are exactly the roots of

t2−s t+p=0.t^2 - s\,t + p = 0.
Example
Solving a symmetric system

Solve {x+y=5x2+y2=13\begin{cases} x + y = 5 \\ x^2 + y^2 = 13 \end{cases}

Step 1. Let s=x+y=5s = x+y = 5. Use the identity x2+y2=s2−2px^2 + y^2 = s^2 - 2p:

13=52−2p=25−2p  ⇒  2p=12  ⇒  p=xy=613 = 5^2 - 2p = 25 - 2p \;\Rightarrow\; 2p = 12 \;\Rightarrow\; p = xy = 6

Step 2. Now xx and yy are the roots of t2−5t+6=0t^2 - 5t + 6 = 0, which factors as (t−2)(t−3)=0(t-2)(t-3)=0, giving t=2t = 2 or t=3t = 3.

Step 3. So {x,y}={2,3}\{x,y\} = \{2,3\}. The solutions are (2,3)(2,3) and (3,2)(3,2).

Check: 2+3=52 + 3 = 5 ✓   and   22+32=4+9=132^2 + 3^2 = 4 + 9 = 13 ✓

Concept
Linear Programming: Optimizing Over a Region

When a system of linear inequalities defines a feasible region, we often want to maximize (or minimize) a linear objective function such as P=3x+4yP = 3x + 4y. The Corner-Point Theorem says the maximum and minimum of a linear objective always occur at a vertex (corner) of the feasible region.

Steps: (1) Graph the constraints and find the feasible region. (2) Find each corner point (each is the intersection of two boundary lines). (3) Evaluate the objective PP at every corner. (4) The largest value is the maximum; the smallest is the minimum.

Example
Maximizing an objective

Maximize P=40x+30yP = 40x + 30y subject to

{x+y≤82x+y≤10x≥0,  y≥0\begin{cases} x + y \le 8 \\ 2x + y \le 10 \\ x \ge 0,\; y \ge 0 \end{cases}

Find the corners. The boundary lines are x+y=8x+y=8 and 2x+y=102x+y=10, together with the axes.

  • (0,0)(0,0) --- origin.
  • (5,0)(5,0) --- where 2x+y=102x+y=10 meets y=0y=0.
  • (0,8)(0,8) --- where x+y=8x+y=8 meets x=0x=0.
  • (2,6)(2,6) --- solve x+y=8x+y=8 and 2x+y=102x+y=10: subtract to get x=2x=2, then y=6y=6.

Evaluate P=40x+30yP = 40x + 30y at each corner:

Maximum: P=260P = 260 at the corner (2,6)(2,6).

Concept
Nonlinear Preview: a Line Meets a Parabola

Not every system is two lines. When one equation is a parabola (like y=x2y = x^2) and the other is a line, substitution still works, but the result is a quadratic. A line and a parabola can cross in two points, one point (tangent), or none, depending on the discriminant of that quadratic.

Example
Line intersecting a parabola

Solve {y=x2y=x+2\begin{cases} y = x^2 \\ y = x + 2 \end{cases}

Step 1. Substitute y=x2y = x^2 into y=x+2y = x+2:

x2=x+2  ⇒  x2−x−2=0x^2 = x + 2 \;\Rightarrow\; x^2 - x - 2 = 0

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

Step 3. Find each yy from y=x+2y = x+2: if x=2x=2, y=4y=4; if x=−1x=-1, y=1y=1.

Check: y=x2y = x^2 gives 22=42^2 = 4 ✓ and (−1)2=1(-1)^2 = 1 ✓

Two solutions: (2,4)(2,4) and (−1,1)(-1,1).

Concept
A Peek at Determinants: Cramer's Rule

For a 2×22\times2 system

{ax+by=ecx+dy=f\begin{cases} a x + b y = e \\ c x + d y = f \end{cases}

the determinant of the coefficients is

D=∣abcd∣=ad−bc.D = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc.

If D≠0D \ne 0 there is exactly one solution, given by Cramer's Rule:

x=∣ebfd∣D=ed−bfD,y=∣aecf∣D=af−ecD.x = \frac{\begin{vmatrix} e & b \\ f & d \end{vmatrix}}{D} = \frac{ed - bf}{D}, \qquad y = \frac{\begin{vmatrix} a & e \\ c & f \end{vmatrix}}{D} = \frac{af - ec}{D}.

Each numerator replaces one coefficient column with the constants ef\begin{smallmatrix} e \\ f \end{smallmatrix}. If D=0D = 0, the system has no unique solution (either none or infinitely many).

Example
Cramer's Rule in action

Solve {2x+3y=74x−y=7\begin{cases} 2x + 3y = 7 \\ 4x - y = 7 \end{cases} using determinants.

Here a=2,  b=3,  e=7a=2,\; b=3,\; e=7 and c=4,  d=−1,  f=7c=4,\; d=-1,\; f=7.

Main determinant:   D=∣234−1∣=(2)(−1)−(3)(4)=−2−12=−14.\;D = \begin{vmatrix} 2 & 3 \\ 4 & -1 \end{vmatrix} = (2)(-1) - (3)(4) = -2 - 12 = -14.

Solve for xx:   x=∣737−1∣D=(7)(−1)−(3)(7)−14=−28−14=2.\;x = \dfrac{\begin{vmatrix} 7 & 3 \\ 7 & -1 \end{vmatrix}}{D} = \dfrac{(7)(-1) - (3)(7)}{-14} = \dfrac{-28}{-14} = 2.

Solve for yy:   y=∣2747∣D=(2)(7)−(7)(4)−14=−14−14=1.\;y = \dfrac{\begin{vmatrix} 2 & 7 \\ 4 & 7 \end{vmatrix}}{D} = \dfrac{(2)(7) - (7)(4)}{-14} = \dfrac{-14}{-14} = 1.

Solution: (2,1)(2,1), matching the elimination result earlier.

Tip

Big picture: Every method here rests on the same idea --- reduce a hard system to a simpler one. Three-variable elimination shrinks to a 2×22\times2; symmetric systems shrink to a single quadratic in tt; Cramer's Rule packages the answer as a ratio of determinants. Whenever a determinant is 00 (or ratios all match), watch for the “no solution / infinitely many” special cases.

Formulas, Proofs & Tips

Tip
Solving systems
Substitution or elimination;parallel⇒no solution;same line⇒infinitely many\text{Substitution or elimination};\quad \text{parallel} \Rightarrow \text{no solution};\quad \text{same line} \Rightarrow \text{infinitely many}

What it means. A solution is a point on both lines, so the answer count is how many times they meet.

Example. x+y=5x+y=5 and x−y=1x-y=1: adding gives 2x=62x=6, so x=3x=3, y=2y=2.

Why it works. Elimination adds a multiple of one equation to the other. Since both sides are equal quantities, the new equation still holds for every common solution — but one variable is gone.

Tip. If the variables all vanish and leave a false statement (0=50=5), there is no solution; a true one (0=00=0) means infinitely many.