Solving Simultaneous Equations: 5a2b -8, a - 3b -5
Introduction
Simultaneous equations are a fundamental aspect of algebra, often appearing in various scientific and engineering fields. In this article, we will delve into solving the simultaneous equations: 5a2b -8 and a - 3b -5. We will explore several methods to find the values of a and b, ensuring a clear and concise solution.Method 1: Combining like terms
Let's start with the given equations:
5a2b -8
a - 3b -5
First, we multiply the second equation by 5 to align the coefficients of b with the first equation:
5a - 15b -25
Now, subtract the first equation from this new equation:
(5a - 15b) - (5a2b) -25 - (-8)
This simplifies to:
5a - 15b - 5a2b -17
Given that 5a2b -8, we substitute:
5a - 15b 8 -17
Simplifying further:
5a - 15b -25
We can solve for b.
-15b -17
b 1
Now substitute b 1 into the second original equation:
a - 3(1) -5
a - 3 -5
a -2
Thus, the solution is (a, b) (-2, 1).
Method 2: Elimination Method
We revisit the equations for a clearer approach:
5a2b -8
a - 3b -5
Multiply the second equation by 5:
5a - 15b -25
Subtract the first equation:
5a - 15b - 5a2b -25 - (-8)
5a - 15b - 5a2b -17
Given 5a2b -8, substitute:
5a - 15b 8 -17
Solve for b:
-15b -25
b 1
Substitute b 1 into the second equation:
a - 3(1) -5
a -2
Therefore, the solution is (a, b) (-2, 1).
Method 3: Substitution Method
Using the second equation directly:
a - 3b -5
Solve for a:
a 3b - 5
Substitute into the first equation:
5(3b - 5)2b -8
15b - 252b -8
15b - 50b -8
-34b -17
b -17 / -34 1/2
Since this does not match our first approaches, double-check the arithmetic. Correctly we have:
15b - 25 -8
17b 17
b 1
Solving for a from a 3b - 5:
a 3(1) - 5
a -2
Thus, the solution is (a, b) (-2, 1).