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Linear Algebra 5
- Module
- Algebra (MA10510)
- Institution
- Aberystwyth University (UWA)
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1. Example 1:Solve for \(x\): \(5x + 2y = 12\)
\(3x - 4y = 8\)
**Step-by-step workings:**
We have a system of two linear equations. Let's use the method of substitution to solve the system.
From the first equation, isolate \(x\) in terms of \(y\):
\(5x = 12 - 2y\)
\(x = \frac{12 - 2y}{5}\)
Substitute this expression for \(x\) into the second equation:
\(3\left(\frac{12 - 2y}{5}\right) - 4y = 8\)
Solve for \(y\):
\(12 - 2y - 4y = 40\)
\(-6y = 28\)
\(y = -\frac{14}{3}\)
Now, substitute the value of \(y\) back into the expression for \(x\):
\(x = \frac{12 - 2\left(-\frac{14}{3}\right)}{5}\)
\(x = \frac{56}{15}\)
**Answer:** \(x = \frac{56}{15}\), \(y = -\frac{14}{3}\)
2. Example 2:
Solve for \(x\) and \(y\): \(2x + 3y = 7\)
\(4x - 5y = -3\)
**Step-by-step workings:**
We have a system of two linear equations. Let's use the method of elimination to solve the system.
Multiply the first equation by 5 and the second equation by 3 to eliminate \(y\):
\(10x + 15y = 35\)
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