Hello, and welcome to today's mathematics lesson. We are going to explore simultaneous linear equations, including how to solve word problems using these powerful techniques. By the end of this lesson, you will understand what simultaneous equations are, master three different methods to solve them, and learn how to apply them to real-life situations.
Let us begin with the foundation. A linear equation in two variables takes the form ax + by + c = 0, where a, b, and c are real constants, while x and y are variables, each with degree one. Now, when we have two such equations involving the same two variables, we call them simultaneous linear equations.
Here is the key idea: a solution to simultaneous equations is a pair of values for x and y that satisfies both equations at the same time. For example, if we have 2x - y = 1 and 3x + y = 14, the values x = 3 and y = 5 work in both equations. Substituting, we get 2 × 3 - 5 = 1 and 3 × 3 + 5 = 14. Since both equations are satisfied, this is our solution.
We will now learn three algebraic methods to solve simultaneous equations. The first is elimination by substitution.
The method of elimination by substitution follows three clear steps. First, express one variable in terms of the other from either equation. Second, substitute this expression into the other equation and solve for the remaining variable. Third, substitute this value back to find the first variable.
Let us work through an example: solve x + y = 7 and 3x - 2y = 11. From the first equation, we get y = 7 - x. Substituting into the second equation: 3x - 2(7 - x) = 11. This simplifies to 3x - 14 + 2x = 11, giving 5x = 25, so x = 5. Finally, y = 7 - 5 = 2. Our solution is x = 5 and y = 2.
Notice that we could equally have started by expressing x in terms of y. The result would be identical. This flexibility is one of the strengths of the substitution method.
The second method is elimination by equating coefficients.
This method is particularly efficient when coefficients are easily matched. The steps are: multiply one or both equations so that the coefficients of one variable become numerically equal. Then add or subtract the equations to eliminate that variable, solve for the remaining variable, and substitute back.
Consider 3x - 4y = 10 and 5x - 3y = 24. Multiply the first equation by 5 and the second by 3, giving 15x - 20y = 50 and 15x - 9y = 72. Subtracting: -11y = -22, so y = 2. Substituting back, 3x - 8 = 10, giving x = 6.
A special case occurs when coefficients are symmetrically arranged. For equations like 65x - 33y = 97 and 33x - 65y = 1, adding gives 98x - 98y = 98, so x - y = 1. Subtracting gives 32x + 32y = 96, so x + y = 3. These simpler equations quickly yield x = 2 and y = 1.
The third method is cross-multiplication, which provides a direct formulaic approach.
When equations are written as a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, the solution is given by:
The formulas are:
x = (b₁c₂ - b₂c₁)/(a₁b₂ - a₂b₁)
and
y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁) y = (c₁a₂ - c₂a₁)/(a₁b₂ - a₂b₁)
The denominator a₁b₂ - a₂b₁ is the same for both variables. A helpful pattern: write the coefficients in columns as b₁, c₁, a₁, b₁ and b₂, c₂, a₂, b₂, then cross-multiply and subtract appropriately.
For example, solve 4x - 7y + 28 = 0 and 7x - 5y - 9 = 0. We identify a₁ = 4, b₁ = -7, c₁ = 28 and a₂ = 7, b₂ = -5, c₂ = -9. Computing: x = ((-7)(-9) - (28)(-5))/((4)(-5) - (7)(-7)) which simplifies to x = 203/29, or simply x = 7. Similarly, y = (28 × 7 - (-9) × 4)/29 which equals y = 232/29 = 8. Similarly, y = (28 × 7 - (-9) × 4)/29 which equals y = 232/29 = 8.
Some equations appear complex but can be reduced to linear form.
When variables appear in denominators, like 7/x + 8/y = 2 and 2/x + 13/y = 22, we use substitution. Let a = 1/x and b = 1/y. The equations become 7a + 8b = 2 and 2a + 13b = 22, which we solve using elimination by equating coefficients. Multiplying the first equation by 2 and the second by 7, then subtracting, gives -75b = -150, so b = 2. Substituting back, 2a + 26 = 22, giving a = -2. Since a = 1/x and b = 1/y, we recover x = -1/2 and y = 1/2.
For equations with expressions like x - y and x + y in denominators, substitute a = 1/(x - y) and b = 1/(x + y), or let a = x - y and b = x + y directly. This transformation reveals the underlying linear structure.
Similarly, for equations with expressions like x - y and x + y in denominators, substitute a = 1/(x - y) and b = 1/(x + y), or let a = x - y and b = x + y directly. This transformation reveals the underlying linear structure.
Now we turn to word problems, where simultaneous equations become powerful tools for real-world situations.
For problems involving two numbers, define your variables clearly. If the sum is 12 and difference is 2, let the numbers be x and y. Then x + y = 12 and x - y = 2. Solving gives 7 and 5.
For fraction problems, let the fraction be x/y. If decreasing the numerator by 1 gives two-thirds, then (x - 1)/y = 2/3. If increasing the denominator by 5 gives one-half, then x/(y + 5) = 1/2. Cross-multiplying yields linear equations to solve.
Two-digit numbers require special attention. If the tens digit is x and units digit is y, the number is 10x + y. Reversed, it becomes 10y + x. Always consider: does reversing increase or decrease the value? This tells you whether y - x = 3 or x - y = 3.
Age problems track how relationships change over time. Seven years hence means adding 7 to current ages. Five years ago means subtracting 5. Set up equations for the given time periods and solve.
For cost price and selling price problems, remember: selling price equals cost price plus profit. If profit is 25 percent, selling price equals cost price plus 25 percent of cost price, which is 5/4 times the cost price.
Work problems use the concept that if a person completes work in x days, their one day's work is 1/x. Combined work adds these rates.
Let us recap the essential points from today's lesson.
First, simultaneous linear equations involve two equations with the same two variables, and we seek values that satisfy both equations simultaneously.
Second, the three solution methods are: elimination by substitution, where we express one variable and substitute; elimination by equating coefficients, where we match and eliminate; and cross-multiplication, which uses a direct formula.
Third, equations with variables in denominators can be transformed using substitutions like a = 1/x or a = 1/(x - y).
Fourth, for word problems, carefully define variables, translate conditions into equations, and verify that your answer makes sense in the original context.
Fifth, for two-digit numbers, remember the structure 10x + y and consider carefully whether the reversed number is larger or smaller.
Sixth, always check your solution by substituting back into both original equations.
You have now learned powerful techniques for solving simultaneous linear equations and applying them to diverse problems. Practice these methods until they feel natural, and you will find them invaluable throughout your mathematical journey. Thank you for your attention, and keep exploring the beautiful patterns of mathematics.