We are asked to show that the system of equations 2x-y=-4 and 8x-4y=8 has no solutions.

A system of linear equations is said to be consistent if it has solutions. If there are two linear equation in two unknowns, a system with a unique solution is said to be consistent and independent, whereas such a system with multiple solutions is consistent and dependent. A system with no solutions is inconsistent.

A linear system of two equations describes two lines. In a plane the lines could intersect at a point (consistent and independent), be parallel (inconsistent), or be the same line (consistent and dependent.)

**I.** We can solve the system using substitution:

2x-y=-4 ==> y=2x+4. Substitute the expression for y into the equation 8x-4y=8 to get:

8x-4(2x+4)=8

8x-8x-16=8

-16=8

There are no values of x that make -16=8 so there are no solutions to the system.

**II.** We can use linear combinations:

2x-y=-4

8x-4y=8

------------- Multiply the first equation by -4 (to eliminate y)

-8x+4y=16

8x-4y=8

-------------- Add these equations

0=24 Again no value for x makes this a true statement.

**III.** Rewrite each equation in slope-intercept form:

2x-y=-4 ==> y=2x+4

8x-4y=8 ==> 4y=8x-8 ==> y=2x-2

Both lines have the same slope (m=2) but different y-intercepts. Thus the lines are parallel and do not have an intersection. There is no solution.

**IV.** Graph both lines

**Further Reading**

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