Tenth Grade Mathematics: Equations Reducible to Linear Form
In many practical scientific and mathematical problems, the equations presented are not initially linear because the variables appear in denominators, under square roots, or as products of variables. However, these complex expressions often exhibit an internal symmetry where identical non-linear terms repeat throughout both equations. By defining auxiliary variables to represent these repeating expressions, we can transform (or reduce) the non-linear system into a standard linear system, solve it using standard methods, and then invert the substitution to find the original unknowns.
The Principle of Auxiliary Variable Substitution
The technique of algebraic reduction involves recognizing a common structural motif:
Transformation Paradigm:
Given non-linear terms such as 1/x and 1/y, introduce new linear variables:
u = 1 / x and v = 1 / y
This transforms reciprocal terms into linear monomials:
Original Non-linear System: Transformed Linear System:
a1 * (1/x) + b1 * (1/y) = c1 ===> a1*u + b1*v = c1
a2 * (1/x) + b2 * (1/y) = c2 ===> a2*u + b2*v = c2
Once the linear system in u and v is solved, the original variables are retrieved by taking reciprocals: x = 1/u and y = 1/v.
Common Reducible Patterns
Several recurring forms appear throughout tenth-grade mathematics:
+-----------------------------+-----------------------------+-------------------------------+
| Original Non-linear Form | Recommended Substitution | Resulting Form |
+-----------------------------+-----------------------------+-------------------------------+
| a/x + b/y = c | u = 1/x, v = 1/y | a*u + b*v = c |
+-----------------------------+-----------------------------+-------------------------------+
| a/(x - 1) + b/(y - 2) = c | u = 1/(x - 1), | a*u + b*v = c |
| | v = 1/(y - 2) | |
+-----------------------------+-----------------------------+-------------------------------+
| a/(x + y) + b/(x - y) = c | u = 1/(x + y), | a*u + b*v = c |
| | v = 1/(x - y) | |
+-----------------------------+-----------------------------+-------------------------------+
| a*x + b*y = c*x*y | Divide both sides by x*y: | b*(1/x) + a*(1/y) = c |
| | a/y + b/x = c | Linear in 1/x, 1/y |
+-----------------------------+-----------------------------+-------------------------------+
Detailed Worked Examples
Example 1: Sum and Difference in Denominators
Solve the following system for real values of x and y:
10 / (x + y) + 2 / (x - y) = 4
15 / (x + y) - 5 / (x - y) = -2
Step 1: Define auxiliary variables:
Let u = 1 / (x + y) and v = 1 / (x - y)
The system becomes:
10*u + 2*v = 4 ===> Divide by 2: 5*u + v = 2 --- (Equation 1)
15*u - 5*v = -2 --- (Equation 2)
Step 2: Solve for u and v using elimination. From Equation 1, v = 2 - 5*u. Substitute into Equation 2:
15*u - 5*(2 - 5*u) = -2
15*u - 10 + 25*u = -2
40*u = 8
u = 8 / 40 = 1 / 5
Now substitute u = 1/5 into v = 2 - 5*u:
v = 2 - 5*(1 / 5) = 2 - 1 = 1
Step 3: Revert to the original variables:
u = 1 / 5 ===> 1 / (x + y) = 1 / 5 ===> x + y = 5 --- (Equation A)
v = 1 ===> 1 / (x - y) = 1 ===> x - y = 1 --- (Equation B)
Step 4: Solve the elementary linear system for x and y:
Add Equation A and Equation B:
(x + y) + (x - y) = 5 + 1 ===> 2*x = 6 ===> x = 3.
Substitute x = 3 into Equation A:
3 + y = 5 ===> y = 2.
The unique solution is x = 3, y = 2.
Example 2: Systems Involving the Product Term x*y
Solve the system:
6*x + 3*y = 6*x*y
2*x + 4*y = 5*x*y
Assuming x != 0 and y != 0 (since x = 0, y = 0 is a trivial solution), divide both sides of each equation by x*y:
(6*x) / (x*y) + (3*y) / (x*y) = 6 ===> 6/y + 3/x = 6 ===> 3*(1/x) + 6*(1/y) = 6
(2*x) / (x*y) + (4*y) / (x*y) = 5 ===> 2/y + 4/x = 5 ===> 4*(1/x) + 2*(1/y) = 5
Letting u = 1/x and v = 1/y yields the linear system:
3*u + 6*v = 6 ===> u + 2*v = 2 --- (Equation 1)
4*u + 2*v = 5 --- (Equation 2)
Subtract Equation 1 from Equation 2: 3u = 3 ===> u = 1. Substitute u = 1 into Equation 1: 1 + 2v = 2 ===> 2*v = 1 ===> v = 1/2. Invert to find x and y: x = 1/u = 1/1 = 1, and y = 1/v = 1/(1/2) = 2. The non-trivial solution is x = 1, y = 2.
Chapter Exercises
Exercise 1: Solve the system of equations by reducing them to a pair of linear equations: 1/(2x) + 1/(3y) = 2 and 1/(3x) + 1/(2y) = 13/6.
Exercise 2: Solve the reducible system: 5/(x - 1) + 1/(y - 2) = 2 and 6/(x - 1) - 3/(y - 2) = 1.
Exercise 3: Solve the radical-reducible system for positive real numbers x and y: 2/sqrt(x) + 3/sqrt(y) = 2 and 4/sqrt(x) - 9/sqrt(y) = -1.
Solutions and Explanations
Solution 1: Let u = 1/x and v = 1/y. The equations become: (1/2)u + (1/3)v = 2 and (1/3)u + (1/2)v = 13/6. Multiply the first equation by 6 to clear fractions: 3u + 2v = 12. Multiply the second equation by 6: 2u + 3v = 13. To eliminate v, multiply the first equation by 3 and the second by 2: 9u + 6v = 36 and 4u + 6v = 26. Subtract the second from the first: 5u = 10 ==> u = 2. Substitute u = 2 into 3u + 2v = 12: 3(2) + 2v = 12 ==> 6 + 2v = 12 ==> 2*v = 6 ==> v = 3. Now invert: x = 1/u = 1/2, and y = 1/v = 1/3. The solution is x = 1/2, y = 1/3.
Solution 2: Let u = 1/(x - 1) and v = 1/(y - 2). The system becomes: 5u + v = 2 and 6u - 3v = 1. Multiply the first equation by 3: 15u + 3v = 6. Add this to the second equation: (15u + 3v) + (6u - 3v) = 6 + 1 ==> 21u = 7 ==> u = 7/21 = 1/3. Substitute u = 1/3 into 5u + v = 2: 5(1/3) + v = 2 ==> v = 2 - 5/3 = 1/3. Inverting the substitutions: 1/(x - 1) = 1/3 ==> x - 1 = 3 ==> x = 4. 1/(y - 2) = 1/3 ==> y - 2 = 3 ==> y = 5. Thus, the solution is x = 4, y = 5.
Solution 3: Let u = 1/sqrt(x) and v = 1/sqrt(y). The system becomes: 2u + 3v = 2 and 4u - 9v = -1. Multiply the first equation by 3: 6u + 9v = 6. Add this equation to the second equation: (6u + 9v) + (4u - 9v) = 6 + (-1) ==> 10u = 5 ==> u = 5/10 = 1/2. Substitute u = 1/2 into the first equation: 2(1/2) + 3v = 2 ==> 1 + 3v = 2 ==> 3*v = 1 ==> v = 1/3. Now invert to find the square roots: 1/sqrt(x) = 1/2 ==> sqrt(x) = 2 ==> x = 2^2 = 4. 1/sqrt(y) = 1/3 ==> sqrt(y) = 3 ==> y = 3^2 = 9. The unique solution is x = 4, y = 9.