How to Solve a System of Two Linear Equations
Linear equations are encountered in everyday life. For example, if 4 apples and 5 oranges cost $2, and 3 apples and 4 oranges cost $1.55, how much does each apple and each orange cost? This article will show you how to solve any system of two linear equations.
Steps
- 1Determine what the unknowns are and assign them variables. For example, let x = price of one apple, y = price of one orange.
- 2Set up the equations. For example:
- If 4 apples and 5 oranges cost $2, write 4x + 5y = 2. (Eq. 1)
- If 3 apples and 4 oranges cost $1.55, write 3x + 4y = 1.55. (Eq. 2)
- 3Solve
- By the Method of Elimination
In this method we first solve for a variable by eliminating the other variable. Let's solve for x by eliminating y. The steps are
- Multiply Eq. 1 by the coefficient of the other variable (y) in Eq. 2
Multiplying Eq. 1 by 4, we get 16x + 20y = 8 - Multiply Eq. 2 by the coefficient of the other variable (y) in Eq. 1
Multiplying Eq. 2 by 5 we get 15x + 20y = 7.75 - Subtract the two equations
We get x = 0.25. Therefore, price of an apple in the example is $0.25 - To find the value of the other variable (y), plug in the value of x obtained in any of the equations we had.
Plugging in x = 0.25 in Eq. 1 we get,
4(0.25) + 5y = 2
=> 1 + 5y = 2
=> 5y = 2 - 1
=> 5y = 1
=> y = 1/5 = 0.20 Therefore, price of an orange in the example is $0.20
- By the Method of Substitution
Read on the wikiHow How to Solve Simultaneous Equations Using Substitution Method
- By the Method of Elimination
Alternative method by matrix reduction
- Set up the the system of linear equations as a matrix equation.
- Reduce the left-most matrix to the identity matrix using one of these two operations:
- Multiplication of a row on both sides through by a constant;
- Subtraction of a row on both sides by a multiple of another row.
- (Make sure that, when carrying out these operation, you match the columns correctly.) See Figure 1 for an example:
Solving a general system of two linear equations
- Let the equations be ax + by = u (Eq. 1) and cx + dy = v (Eq. 2), where a, b, c, d, u, v are given numbers, and x and y are the unknowns. Assume that a, b are not both zero, and c,d are not both zero.
- To solve for x, multiply the first equation by d and the second equation by b, then subtract, to get (ad - bc)x = du - bv.
- To solve for y, multiply the first equation by c and the second equation by a, then subtract, to get (ad - bc)y = av - cu.
- If ad - bc ≠ 0, divide both equations by ad - bc to get x = (du - bv)/(ad - bc), and y = (av - cu)/(ad - bc). In the example above, a=4, b=5, c=3, d=4, u=2, and v=1.55. Since ad - bc = 1 ≠ 0, plug the numbers in to get x=0.25 and y=0.20.
- If ad - bc = 0, and c ≠ 0, let k = a/c, so a = ck. Substitute into ad = bc to get cdk = cb, or dk = b (sine c ≠ 0). If c = 0, d ≠ 0 by assumption above, so let k = b/d, or b = dk. Substitute into ad = bc to get ad = cdk, or a = ck. If k = 0, then a = b = 0, contrary to assumption. So k ≠ 0. Multiply Eq. 2 by k, to get ckx + dky = kv, or ax + by = kv. If u = kv, then this is the same as Eq. 1, in which case the system has infinitely many solutions (x,y) that satisfy the first equation. If u ≠ kv, then (x,y) satisfying the first equation will not satisfy the second, and the system has no solution.
- In sum:
- If there is no number k such that a = ck and b = dk, then the system of 2 linear equations has a unique solution, given by x = (du - bv)/(ad - bc), and y = (av - cu)/(ad - bc). This is analogous to two straight lines intersecting at one point.
- If there is a number k such that a = ck and b = dk, and u = vk, then the system of 2 linear equations has infinitely many solutions. This is analogous to two straight lines that coincide.
- If there is a number k such that a = ck and b = dk, and u ≠ vk, then the system of 2 linear equations has no solution at all. This is analogous to two straight lines parallel to each other.
Tips
- Matrix reduction method has the advantage of simplicity, and it can also be generalized to solve systems of more than 2 linear equations.
Article Info
Categories: Algebra