10 Easy Simultaneous Equations Examples with Step-by-Step Solutions

Simultaneous equations might sound intimidating, but they are simply a set of equations that share the same variables (like x and y). Solving them allows you to find the one specific point where two mathematical conditions are true at the same time. Whether you are calculating a budget, comparing social media stats, or predicting profit, these skills are vital.

Below are 10 clear examples of linear simultaneous equations, ranging from basic substitution to real-world word problems. Each example follows a “Step-by-Step Solution” format to guide you through the logic.

Part 1: The Fundamentals – Elimination Method

The elimination method focuses on adding or subtracting equations to cancel out (eliminate) one variable.

Example 1: Adding to Eliminate

Solve the system:

x+y=10x−y=4x+yx−y​=10=4​

Step-by-Step Solution:

1. Check the coefficients: Notice the +y in the first equation and the -y in the second. If we add these, the y variables will cancel out instantly because $y+(-y)=0$.

2. Add the equations:

   $$(x+x)+(y-y)=10+4$$$$2x=14$$

3. Solve for x:

   x=142=7x=214​=7

4. Substitute back: Plug $x=7$ into the first equation.

   $$7+y=10$$$$y=3$$

5. Solution: $x=7,y=3$.

Example 2: Subtracting to Eliminate

Solve the system:

3x+2y=162x+2y=123x+2y2x+2y​=16=12​

Step-by-Step Solution:

1. Observe: Both equations have $+2y$. If we subtract one equation from the other, the y terms cancel.

2. Subtract Equation 2 from Equation 1:

   $$(3x-2x)+(2y-2y)=16-12$$$$x=4$$

3. Substitute $x=4$ into $2x+2y=12$:

   $$2(4)+2y=12$$$$8+2y=12⟹2y=4⟹y=2$$

4. Solution: $x=4,y=2$.

Example 3: Multiplying One Equation (The Times Table Strategy)

Sometimes the variables don’t line up perfectly. You must multiply one equation by a number to make them match.

Solve the system:

2x+5y=96x−4y=82x+5y6x−4y​=9=8​

Step-by-Step Solution:

1. The Plan: Look at the x terms (2x and 6x). If we multiply the entire first equation by 3, we get 6x, which will match the second equation.

2. Multiply the whole equation: $3\times (2x+5y=9)$ becomes $6x+15y=27$.

3. Subtract the equations: Now we have:

   6x+15y=276x−4y=86x+15y6x−4y​=27=8​

   Subtract to eliminate 6x: $(15y-(-4y))=27-8$ => $19y=19$ => $y=1$.

4. Solve for x: Substitute $y=1$ into $2x+5(1)=9$ => $2x=4$ => $x=2$.

5. Solution: $x=2,y=1$.

Example 4: Multiplying Both Equations

Solve the system:

3x+5y=144x+3y=153x+5y4x+3y​=14=15​

Step-by-Step Solution:

1. Match Coefficients: We need to make the x terms match. The lowest common multiple of 3 and 4 is 12.

   • Multiply Equation 1 by 4: $12x+20y=56$

   • Multiply Equation 2 by 3: $12x+9y=45$.

2. Subtract:

   $$(12x-12x)+(20y-9y)=56-45$$$$11y=11⟹y=1$$

3. Substitute: $y=1$ into $3x+5(1)=14$ => $3x=9$ => $x=3$.

4. Solution: $x=3,y=1$.

Part 2: Real-World & Social Media Scenarios

Simultaneous equations are everywhere in digital marketing and social media analytics. Let’s look at two examples involving social media links and follower counts.

Example 5: The Social Media Follower Count

Problem: Abby uses two social media sites. On Site A, she has 52 more followers than Site B. Combined, she has 700 followers total. How many followers does she have on each site? Note: This scenario uses data similar to common Instagram and X (Twitter) analytics queries.

Step-by-Step Solution:

1. Define Variables:
   Let $A$ = followers on Site A.
   Let $B$ = followers on Site B.

2. Form Equations:

   • Info 1: $A=B+52$ (52 more on A)

   • Info 2: $A+B=700$ (Total followers)

3. Substitution Method: Since we know $A=B+52$, plug it into the total equation.

   $$(B+52)+B=700$$$$2B+52=700$$$$2B=648⟹B=324$$

4. Find A: $A=324+52=376$.

5. Conclusion: Site A has 376 followers; Site B has 324 followers.

Example 6: Comparing Engagement Rates

Problem: On LinkedIn, a professional posted two updates. Update 1 received 6 likes and 19 shares, generating 240 click-throughs. Update 2 received 11 likes and 14 shares, generating 190 click-throughs. Assuming every like and share generates a specific number of clicks, how many clicks does one “like” generate?
(Derived from coefficient strategies)

Step-by-Step Solution:

1. Define Variables:
   Let $x$ = clicks generated by 1 Like.
   Let $y$ = clicks generated by 1 Share.

2. Form Equations:

   $$6x+19y=240$$$$11x+14y=190$$

3. Elimination Strategy: We need to eliminate $y$. Multiply the first by 14 and the second by 19:

   • Eq 1 ($\times 14$): $84x+266y=3360$

   • Eq 2 ($\times 19$): $209x+266y=3610$

4. Subtract: $(209x-84x)+(266y-266y)=3610-3360$

   $$125x=250⟹x=2$$

5. Conclusion: One “like” generates 2 clicks.

Part 3: The Substitution Method

Sometimes, solving for one variable and plugging it into the other equation is faster.

Example 7: Easy Substitution

Solve the system:

y=2x+13x+2y=9y3x+2y​=2x+1=9​

Step-by-Step Solution:

1. Notice: Equation 1 already tells us what y equals.

2. Substitute: Replace y in Equation 2 with $2x+1$.

   $$3x+2(2x+1)=9$$$$3x+4x+2=9$$$$7x=7⟹x=1$$

3. Find y: $y=2(1)+1=3$.

4. Solution: $x=1,y=3$.

Example 8: Rearrange and Substitute

Solve the system:

2x+y=7x−y=22x+yx−y​=7=2​

Step-by-Step Solution:

1. Isolate: Using the second equation ($x-y=2$), solve for $x$: $x=y+2$.

2. Substitute into $2x+y=7$:

   $$2(y+2)+y=7$$$$2y+4+y=7$$$$3y=3⟹y=1$$

3. Find x: $x=1+2=3$.

4. Solution: $x=3,y=1$.

Part 4: Word Problems & Practical Scenarios

Example 9: The Cinema Ticket Problem

Problem: At a cinema, a family bought 2 drinks and 4 popcorn bags for £15. Another family bought 3 drinks and 2 popcorn bags for £10.50. Find the cost of one drink.

Step-by-Step Solution:

1. Variables:
   Let $d$ = cost of a drink.
   Let $p$ = cost of popcorn.

2. Equations:

   $$2d+4p=15.00$$$$3d+2p=10.50$$

3. Eliminate $p$: Multiply the second equation by 2 to match the $p$ coefficients.

   • $\times 2$: $6d+4p=21.00$

4. Subtract the first equation from this new equation:

   $$(6d-2d)+(4p-4p)=21.00-15.00$$$$4d=6.00⟹d=1.50$$

5. Substitute: $2(1.50)+4p=15$ => $3+4p=15$ => $4p=12$ => $p=3$.

6. Answer: A drink costs £1.50, and popcorn costs £3.00.

Example 10: The Rectangle Perimeter

Problem: The perimeter of a rectangle is 50 units. The area is 100 square units. Find the side lengths.

Step-by-Step Solution:

1. Variables:
   Let $l$ = length, $w$ = width.

2. Formulas:

   • Perimeter: $2l+2w=50$ (Simplify: $l+w=25$)

   • Area: $l\times w=100$

3. Substitution: From the perimeter, $w=25-l$. Plug into the area:

   $$l\times (25-l)=100$$25l−l2=10025l−l2=1000=l2−25l+1000=l2−25l+100$$(l-5)(l-20)=0$$

4. Solutions: $l=5$ or $l=20$.

   • If $l=5$, $w=20$.

   • If $l=20$, $w=5$.

5. Answer: The sides are 20 units and 5 units.

Part 5: FAQ – Simultaneous Equations

Here are the top five questions students ask when studying this topic.

Q1: What exactly are simultaneous equations?

Simultaneous equations are two or more equations that share variables. You need to find a value for x That works for all equations at the same time. ime If you only solve one equation, there are infinite solutions; the second equation narrows it down to just one (or a few).

Q2: How do I know which method to use (Elimination vs. Substitution)?

• Use Elimination when both equations are in the standard form $Ax+By=C$ and the coefficients are easy to multiply to match.

• Use Substitution when one variable is already isolated (e.g., $y=3x+2$) or can be isolated very easily (e.g., $x+y=10$).

Q3: I multiplied the equations, but my answer doesn’t work. Why?

This is usually a sign of an error! Remember: If you are adding to eliminate, be careful with negatives. If you are subtracting, make sure you subtract every term.

• Pro Tip: If you get a strange decimal, plug it back into the original equations to check your work

Q4: Can I solve simultaneous equations on a graph?

Yes. The solution to a pair of linear equations is the point where the two lines cross ($x,y$). However, graphical methods can be inaccurate if the answer is a fraction or decimal, which is why algebra (the methods above) is preferred for accuracy.

Q5: How do I handle equations with three variables (x, y, z)?

The logic is the same, but it takes longer. You use the equations times two to eliminate one variable, leaving you with two equations and two unknowns. You then solve those two using the methods above.

Part 6: Why This Matters & Social Learning

Solving these equations is a fundamental algebraic skill used in business analytics, engineering, and computer science. To see how others approach these problems or to share your own solutions, connect with study communities on Instagram (look for the hashtag #MathHelp) or join discussion groups on LinkedIn. By mastering these 10 examples, you have built the foundation to solve complex systems of equations with confidence. Keep practicing the times tables for coefficients, and the logic will become second nature.