# Systems of Linear Equations

4.3. A Short Quiz to Test Yourself

Now try a few for yourselves. After you solve the problem , click on the green link

Quiz Solve each of the fol lowing systems using any method you choose .

1.

(4,−1)
(−9, 3)
(−8, 5)
(2, 2)
No solution
{(x, y) | x + 3y = 0}

2.

(4,−1)
(−9, 3)
(−8, 5)
(2, 2)
No Solution
{(x, y) | x − 5y = −3}

3.

(4,−1)
(−9, 3)
(−8, 5)
(2, 2)
No solution
{(x, y) | 2x − y = 9}

4.

(4,−1)
(−9, 3)
(−8, 5)
(2, 2)
No solution
{(x, y) | −x + 3y = 4}

5.

(4,−1)
(−9, 3)
(−8, 5)
(2, 2)
No solution
{(x, y) | 2x+3y = −1}

How did you do?

Solutions to Examples

Example 2.1(a): We substitute the point (−5, 3) into the equation to get

You can see that the first equation is satisfied, but the second equation is not.

Conclusion: The point (−5, 3) does not satisfy the system.

Example 2.1(b): We substitute the point (−3, 4) into the equation to get

You can see that the both equations are satisfied.

Conclusion: The point (−3, 4) does satisfy the system.

Example 4.1(a): We solve for y in the first equation of   to get
We now substitute into the second equation:

second equation
substitution
combine
and solve

Now substituting x = 6 into the we get y = −2.

Solution: These are consistent equations with a unique solution of (6,−2) , or
the solution set is { (6,−2) } .

Example 4.1(b): We solve for y in the first equation of to
get y = 1− 3x, and substitute into the second equation:

−6x − 2y = −4 second equation
−6x − 2(1 − 3x) = −4 substitution
−6x − 2 + 6x = −5 combine
−2 = −5 simplify (2)

We see that in equation (2) we get a “false” equation. This means that there is
no solution to the system.

Solution: This is an inconsistent system of equations, there is no solution to this
system. The solution set is .

Example 4.1(c): We solve for x in the first equation in to get
x = 6− 3y. Substitute into the second equation:

second equation
substitution
factor out 3
combine
deal with the minus sign , correctly
simplify again (3)

Equation (3) is typical of a dependent system of equations. The equation is always
true regardless of the value of x . Dependent systems means essentially that there
are two equations , but each equation is describing the same line .

Solution: This is an consistent system of dependent equations. Any point that
satisfies one of the equations automatically satisfies the the other equation. Therefore,
the solution set is the set of all points on the satisfy either of the two equations.
We can write { (x, y) | x + 3y = 6}

Example 4.4(a): To solve

we simply add the two equations together to get 3y = 9, hence, y = 3. From the
first equation, we see that x = −5 − x = −5 −3 = −8.

Conclusion: This is a consistent system of equations with a unique solution of
(−8, 3) .

Example 4.4(b): As discussed earlier, we multiply the first equation by −3:

given

multiply by −3
solve for y

From the first equation x = −5 − 2y = −5 − 2(−8) = −5 + 16 = 11

Conclusions: This is a consistent system of equations with a unique solution of
(11,−8) .

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