How to Solve Algebra Math Problems? | Simultaneous Equation Puzzles

Solving simultaneous equation puzzles can take some investigation skills, a few analytical skills, a little deductive reasoning, a bit of patience, and sometimes, a tiny bit of luck, too. To solve a group of simultaneous equations, first try to find two or more equations that may be combined and simplified (a simplified equation has a reduced number of unknowns). Then try to repeat this process with more equations until you’ve derived an equation that has only one or two unknowns. After you’ve done this, you’ll either have a numeric solution for one of the unknowns, or you’ll have an equation that expresses an unknown in terms of just one other unknown. At this point, you can begin substituting your answer for this unknown into other equations, until you’ve found answers for all the other unknowns.

Below is a sample simultaneous equation puzzle we’ll use to demonstrate some useful techniques for solving simultaneous equations.

Sample Puzzle:

Find the 6-digit number ABCDEF by solving the following equations:

eq.1) A + B + C + D + E + F = 29 eq.2) E = C + D eq.3) F – A = C eq.4) D + F = A + B + C eq.5) B – C + D = F – E eq.6) F = B × D

A, B, C, D, E, and F each represent a one-digit non-negative integer.

There are only a few techniques you’ll need to know to combine and simplify equations. With these techniques, you’ll be able to find the solution to almost any simultaneous equation puzzle on SimEqPuzzles.com. These are:

1) the equivalent substitution method
2) the identical modification method
3) the trial-and-error method

Although these techniques are different from one another, the first two follow the same simple rule:

Any change to any part of any equation may be made, provided that both sides of the equation remain equal.

1) The equivalent substitution method The simplest and most often used technique is probably the equivalent substitution method. With this technique, you replace or substitute a mathematical expression in one equation with an equivalent but different expression from another equation. The replacement expression must be equivalent to the replaced expression. Using this technique (you may need to use it several times) will almost always help you find an equation that expresses an unknown in terms of just one other unknown. This technique is demonstrated by the equations below.

Substitute F – A for C (from eq.3) in eq.4: D + F = A + B + F – A which becomes eq.3a) D + F = B + F Substitute (C + D) for E (from eq.2) in eq.5: B – C + D = F – (C + D) which becomes eq.5a) B – C + D = F – C – D

2) The identical modification method The second technique may be called the identical modification method. With this technique, you can modify an equation by applying the same mathematical operation and expression to both sides — i.e., you make the same change to each side of the equation. You may use × or ÷ to multiply or divide each side by a constant, or you may use + or – to add or subtract one or more unknowns to or from each side. Any of these may be used, as long as both sides of the equation remain equal. Used together, the equivalent substitution and identical modification methods will almost always lead to finding an equation that expresses an unknown in terms of just one other unknown. This identical modification method is demonstrated by the equations below.

Subtract F from both sides of eq.3a: D + F – F = B + F – F which simplifies to D = B Add C + D to both sides of eq.5a: B – C + D + C + D = F – C – D + C + D which becomes eq.5b) B + 2×D = F

1 (again) The equivalent substitution method You can use the equivalent substitution and the identical modification methods in any order, and as much as you need to. To help solve the sample puzzle, the equivalent substitution technique is used again as demonstrated by the equations below.

Because D = B, replace D with B in eq.5b: B + 2×B = F which means eq.5c) 3×B = F Now substitute B for D, and 3×B for F in eq.6: 3×B = B × B making B = 0 or B = 3

3) The trial-and-error method A third technique that may be used is the trial-and-error method. Trial-and-error is frequently the most time-consuming technique, so you would usually use this technique only as a last resort. But sometimes it’s the best way to find a solution to one or more unknowns in a set of simultaneous equations. You’d frequently use trial-and-error when there are more unknowns than equations, or when an equation has a non-linear math operation (such as multiplication, division, exponentiation, logarithm, or trigonometry function). This technique is demonstrated by the equations below.

Check: B = 0 … Substitute 0 for B in eq.6: F = 0 × D which would make F = 0 In eq.3, replace F with 0: 0 – A = C which would mean -A = C Since A and C must be non-negative integers, then A = 0 and C = 0 Substitute 0 for A, B, C, D, and F in eq.1: 0 + 0 + 0 + 0 + E + 0 = 29 which would make E = 29 Since E must be a one-digit integer, then E ≠ 29 which means B ≠ 0 and therefore makes B = 3 and D = 3

There are also some other techniques that can be used to solve simultaneous equation problems. But, with very few exceptions, the techniques described here are the only ones you’ll need to solve the big majority of puzzles on SimEqPuzzles.com. If you are interested in learning more about solving simultaneous equations, search the Internet for the keywords “simultaneous equations”. Your search results will return several websites that give examples of simultaneous equations, and tips on how to solve them.

Here’s the rest of the solution to the sample puzzle. Only the equivalent substitution and the identical modification methods are needed to finish the puzzle’s solution.

In eq.5c, replace B with 3: 3×3 = F which means 9 = F Substitute 3 for B and D, and 9 for F in eq.4: 3 + 9 = A + 3 + C Subtract both 3 and A from each side of the equation above: 3 + 9 – 3 – A = A + 3 + C – 3 – A which becomes eq.4a) 9 – A = C Substitute 9 – A for C (from eq.4a), and 3 for D in eq.2: E = 9 – A + 3 which is the same as eq.2a) E = 12 – A Substitute 3 for B and D, 9 – A for C (from eq.4a), 12 – A for E (from eq.2a), and 9 for F in eq.1: A + 3 + 9 – A + 3 + 12 – A + 9 = 29 which becomes -A + 36 = 29 Add (A – 29) to both sides of the above equation: -A + 36 + (A – 29) = 29 + (A – 29) which makes 7 = A In eq.4a, replace A with 7: 9 – 7 = C which makes 2 = C Substitute 7 for A in eq.2a: E = 12 – 7 = 5 making ABCDEF = 732359