Question

1. solve the system of equations

a) $x - y = 3$
$-2x + y = -3$
b) $x + 3y = 9$
$2x - y = 4$
c) $y = 4 - x$
$y = 2x - 3$

Explanation:

Part (a)

Step1: Add the two equations

We have the system:

$$
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$$

Adding the two equations: \((x - y) + (-2x + y) = 3 + (-3)\)
Simplify: \(x - y - 2x + y = 0\) which gives \(-x = 0\)

Step2: Solve for \(x\)

From \(-x = 0\), we get \(x = 0\)

Step3: Substitute \(x = 0\) into the first equation

Substitute \(x = 0\) into \(x - y = 3\): \(0 - y = 3\)
Solve for \(y\): \(y = -3\)

Step1: Solve the first equation for \(x\)

From \(x + 3y = 9\), we get \(x = 9 - 3y\)

Step2: Substitute \(x = 9 - 3y\) into the second equation

Substitute into \(2x - y = 4\): \(2(9 - 3y) - y = 4\)
Expand: \(18 - 6y - y = 4\)
Simplify: \(18 - 7y = 4\)

Step3: Solve for \(y\)

Subtract 18 from both sides: \(-7y = 4 - 18 = -14\)
Divide by -7: \(y = \frac{-14}{-7} = 2\)

Step4: Substitute \(y = 2\) into \(x = 9 - 3y\)

\(x = 9 - 3(2) = 9 - 6 = 3\)

Step1: Set the two equations equal to each other

We have \(y = 4 - x\) and \(y = 2x - 3\)
Set them equal: \(4 - x = 2x - 3\)

Step2: Solve for \(x\)

Add \(x\) to both sides: \(4 = 3x - 3\)
Add 3 to both sides: \(7 = 3x\)
Divide by 3: \(x = \frac{7}{3}\)

Step3: Substitute \(x = \frac{7}{3}\) into \(y = 4 - x\)

\(y = 4 - \frac{7}{3} = \frac{12}{3} - \frac{7}{3} = \frac{5}{3}\)

Answer:

\(x = 0\), \(y = -3\)

Part (b)