Systems of Equations

Solving systems of linear and nonlinear equations using multiple methods

Systems of Equations

What is a System of Equations?

A system of equations is a set of two or more equations with the same variables. A solution is a set of values that satisfies all equations simultaneously.

Types of Systems

Linear System (2 equations, 2 unknowns):

ax + by = c \\ dx + ey = f \end{cases}$$ **Nonlinear System** (may include quadratics, circles, etc.): $$\begin{cases} y = x^2 \\ y = 2x + 3 \end{cases}$$ ## Number of Solutions A system can have: - **One solution**: Lines/curves intersect at exactly one point - **No solution**: Lines/curves are parallel (never intersect) - **Infinitely many solutions**: Lines/curves coincide (same graph) ## Method 1: Substitution **Steps:** 1. Solve one equation for one variable 2. Substitute into the other equation 3. Solve for the remaining variable 4. Back-substitute to find the other variable **Best for:** When one equation is already solved for a variable, or easily can be ### Example $$\begin{cases} y = 2x - 1 \\ 3x + y = 9 \end{cases}$$ Substitute $y = 2x - 1$ into second equation: $$3x + (2x - 1) = 9$$ $$5x - 1 = 9$$ $$5x = 10$$ $$x = 2$$ Then: $y = 2(2) - 1 = 3$ **Solution:** $(2, 3)$ ## Method 2: Elimination (Addition) **Steps:** 1. Multiply equations (if needed) to make coefficients of one variable opposites 2. Add equations to eliminate that variable 3. Solve for the remaining variable 4. Substitute back to find the other variable **Best for:** When coefficients are easy to manipulate ### Example $$\begin{cases} 2x + 3y = 8 \\ x - 3y = 1 \end{cases}$$ Add the equations (the $3y$ terms cancel): $$3x = 9$$ $$x = 3$$ Substitute into second equation: $$3 - 3y = 1$$ $$-3y = -2$$ $$y = \frac{2}{3}$$ **Solution:** $(3, \frac{2}{3})$ ## Method 3: Graphing **Steps:** 1. Graph both equations 2. Find intersection point(s) 3. Verify by substitution **Best for:** Visual understanding, approximate solutions ## Systems with 3 Variables For three variables $(x, y, z)$, you need three equations: $$\begin{cases} ax + by + cz = d \\ ex + fy + gz = h \\ ix + jy + kz = l \end{cases}$$ **Strategy:** Use elimination to reduce to 2 variables, then solve ## Nonlinear Systems For systems involving quadratics, circles, or other curves: - Substitution is usually best - May have 0, 1, 2, or more solutions - Graph to visualize ### Example: Line and Parabola $$\begin{cases} y = x^2 - 4 \\ y = 2x - 1 \end{cases}$$ Substitute: $x^2 - 4 = 2x - 1$ $$x^2 - 2x - 3 = 0$$ $$(x - 3)(x + 1) = 0$$ $$x = 3 \text{ or } x = -1$$ Solutions: $(3, 5)$ and $(-1, -3)$ ## Applications - **Mixture problems**: Combining solutions with different concentrations - **Distance/rate/time**: Two objects moving - **Business**: Supply and demand, break-even analysis - **Geometry**: Finding intersection points

📚 Practice Problems

1Problem 1easy

❓ Question:

Solve the system using substitution: {x+2y=73x−y=5\begin{cases} x + 2y = 7 \\ 3x - y = 5 \end{cases}

💡 Show Solution

Solve using substitution:

Step 1: Solve the first equation for xx: x+2y=7x + 2y = 7 x=7−2yx = 7 - 2y

Step 2: Substitute into the second equation: 3(7−2y)−y=53(7 - 2y) - y = 5 21−6y−y=521 - 6y - y = 5 21−7y=521 - 7y = 5 −7y=−16-7y = -16 y=167y = \frac{16}{7}

Step 3: Find xx: x=7−2(167)=7−327=49−327=177x = 7 - 2\left(\frac{16}{7}\right) = 7 - \frac{32}{7} = \frac{49 - 32}{7} = \frac{17}{7}

Step 4: Verify in both equations: 177+2(167)=17+327=497=7\frac{17}{7} + 2\left(\frac{16}{7}\right) = \frac{17 + 32}{7} = \frac{49}{7} = 7 ✓ 3(177)−167=51−167=357=53\left(\frac{17}{7}\right) - \frac{16}{7} = \frac{51 - 16}{7} = \frac{35}{7} = 5 ✓

Answer: (177,167)\left(\frac{17}{7}, \frac{16}{7}\right)

2Problem 2medium

❓ Question:

Solve using elimination: {2x+5y=133x−2y=−4\begin{cases} 2x + 5y = 13 \\ 3x - 2y = -4 \end{cases}

💡 Show Solution

Solve using elimination:

Step 1: Make coefficients of one variable opposites. Multiply first equation by 2 and second by 5:

4x + 10y = 26 \\ 15x - 10y = -20 \end{cases}$$ Step 2: Add to eliminate $y$: $$19x = 6$$ $$x = \frac{6}{19}$$ Step 3: Substitute into first original equation: $$2\left(\frac{6}{19}\right) + 5y = 13$$ $$\frac{12}{19} + 5y = 13$$ $$5y = 13 - \frac{12}{19} = \frac{247 - 12}{19} = \frac{235}{19}$$ $$y = \frac{235}{95} = \frac{47}{19}$$ Step 4: Verify: $$2\left(\frac{6}{19}\right) + 5\left(\frac{47}{19}\right) = \frac{12 + 235}{19} = \frac{247}{19} = 13$$ ✓ **Answer:** $\left(\frac{6}{19}, \frac{47}{19}\right)$

3Problem 3hard

❓ Question:

Solve the nonlinear system: {x2+y2=25y=x+1\begin{cases} x^2 + y^2 = 25 \\ y = x + 1 \end{cases}

💡 Show Solution

Solve the system (circle and line):

Step 1: Substitute y=x+1y = x + 1 into the first equation: x2+(x+1)2=25x^2 + (x + 1)^2 = 25 x2+x2+2x+1=25x^2 + x^2 + 2x + 1 = 25 2x2+2x+1=252x^2 + 2x + 1 = 25 2x2+2x−24=02x^2 + 2x - 24 = 0 x2+x−12=0x^2 + x - 12 = 0

Step 2: Factor: (x+4)(x−3)=0(x + 4)(x - 3) = 0 x=−4 or x=3x = -4 \text{ or } x = 3

Step 3: Find corresponding yy values:

  • If x=−4x = -4: y=−4+1=−3y = -4 + 1 = -3
  • If x=3x = 3: y=3+1=4y = 3 + 1 = 4

Step 4: Verify both solutions:

For (−4,−3)(-4, -3): (−4)2+(−3)2=16+9=25(-4)^2 + (-3)^2 = 16 + 9 = 25 ✓ −3=−4+1-3 = -4 + 1 ✓

For (3,4)(3, 4): 32+42=9+16=253^2 + 4^2 = 9 + 16 = 25 ✓ 4=3+14 = 3 + 1 ✓

Answer: Two solutions: (−4,−3)(-4, -3) and (3,4)(3, 4)

Interpretation: The line intersects the circle at two points.

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