Question

7 - 10. find the solution for each system of equations below, if a solution exists. if there is not a single solution, explain why not. be sure to check your solution, if possible. homework help a. \\( x + 4y = 2 \\) \\( x + 4y = 10 \\) b. \\( 2x + 4y = -10 \\) \\( x + 2y = -5 \\) c. \\( 3(6 - x) + 2x = 15 \\) for \\( x \\)

Explanation:

Part a

Step1: Analyze the system consistency

Subtract first equation from second:
$(x+4y)-(x+4y)=10-2$
$0=8$

Step2: Interpret the result

The statement $0=8$ is false, so no solution exists.

Part b

Step1: Simplify the first equation

Divide $2x+4y=-10$ by 2:
$x+2y=-5$

Step2: Compare with second equation

This matches the second equation, so infinitely many solutions exist (all $(x,y)$ where $x=-5-2y$ for any real $y$).

Part c

Step1: Expand the left-hand side

$3(6-x)+2x=18-3x+2x$

Step2: Combine like terms

$18-x=15$

Step3: Solve for $x$

Subtract 18 from both sides: $-x=15-18=-3$
Multiply by $-1$: $x=3$

Answer:

a. No solution exists, because subtracting the equations results in the false statement $0=8$, meaning the lines are parallel and never intersect.
b. Infinitely many solutions exist, because the first equation simplifies to exactly the second equation, representing the same line; all solutions are of the form $(x,y)=(-5-2y, y)$ where $y$ is any real number.
c. $x=3$