Question

Solve and explain the system of equations 2x-y=-4 and 8x-4y=8, and demonstrate why they have "no solution."

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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)=88x-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=-48x-4y=8 ------------- Multiply the first equation by -4 (to eliminate y)
-8x+4y=168x-4y=8-------------- Add these equations0=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

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Solve and explain the system of equations 2x-y=-4 and 8x-4y=8, and demonstrate why they have "no solution."

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