Asked by Makayla
Given the system of linear equations.
{ x - y = 5
{ 4x + y = 10
Part A: Use substitution to find the solution to the systems of equations. Include all of your work in your final answer.
Part B: Algebraically verify your answer to Part A. Include all of your work in your final answer.
Part C: Graphically verify your answer to Part A. In your final answer, include the graph.
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Part A — Substitution (showing all work)
System:
x - y = 5
4x + y = 10
From the first equation solve for y:
x - y = 5
=> -y = 5 - x
=> y = x - 5
Substitute y = x - 5 into the second equation:
4x + y = 10
4x + (x - 5) = 10
5x - 5 = 10
5x = 15
x = 3
Now find y:
y = x - 5 = 3 - 5 = -2
Solution: (x, y) = (3, -2).
Part B — Algebraic verification (plug the solution back in)
Check in equation 1: x - y = 5
3 - (-2) = 3 + 2 = 5 ✓
Check in equation 2: 4x + y = 10
4(3) + (-2) = 12 - 2 = 10 ✓
Both equations are satisfied, so (3, -2) is correct. (You can also verify by elimination: adding the two original equations x - y = 5 and 4x + y = 10 gives 5x = 15, x = 3, then y = 3 - 5 = -2.)
Part C — Graphical verification
Rewrite the equations in slope-intercept form:
From x - y = 5 => y = x - 5 (line L1)
From 4x + y = 10 => y = 10 - 4x (line L2)
Choose points for each line:
- L1: y = x - 5
x = -1 → y = -6
x = 0 → y = -5
x = 1 → y = -4
x = 2 → y = -3
x = 3 → y = -2 (intersection)
x = 4 → y = -1
x = 5 → y = 0
- L2: y = 10 - 4x
x = 1 → y = 6
x = 2 → y = 2
x = 3 → y = -2 (intersection)
x = 4 → y = -6
These two lines meet at (3, -2). Here is a simple plotted grid (x from -1 to 6, y from 6 down to -6). Symbols: 1 = a point on L1, 2 = a point on L2, X = intersection (both).
x: -1 0 1 2 3 4 5 6
y= 6 . | 2 . . . . .
y= 5 . | . . . . . .
y= 4 . | . . . . . .
y= 2 . | . 2 . . . .
y= 1 . | . . . . . 1
y= 0 . 0 . . . . 1 .
y=-1 . | . . . 1 . .
y=-2 . | . . X . . .
y=-3 . | . 1 . . . .
y=-4 . | 1 . . . . .
y=-5 . 1 . . . . . .
y=-6 1 | . . . 2 . .
Legend: '|' is the y-axis (x=0), '0' marks (0,0). The X at (3,-2) is the intersection point. This confirms graphically that the two lines meet at (3, -2).
Final answer: (x, y) = (3, -2).
System:
x - y = 5
4x + y = 10
From the first equation solve for y:
x - y = 5
=> -y = 5 - x
=> y = x - 5
Substitute y = x - 5 into the second equation:
4x + y = 10
4x + (x - 5) = 10
5x - 5 = 10
5x = 15
x = 3
Now find y:
y = x - 5 = 3 - 5 = -2
Solution: (x, y) = (3, -2).
Part B — Algebraic verification (plug the solution back in)
Check in equation 1: x - y = 5
3 - (-2) = 3 + 2 = 5 ✓
Check in equation 2: 4x + y = 10
4(3) + (-2) = 12 - 2 = 10 ✓
Both equations are satisfied, so (3, -2) is correct. (You can also verify by elimination: adding the two original equations x - y = 5 and 4x + y = 10 gives 5x = 15, x = 3, then y = 3 - 5 = -2.)
Part C — Graphical verification
Rewrite the equations in slope-intercept form:
From x - y = 5 => y = x - 5 (line L1)
From 4x + y = 10 => y = 10 - 4x (line L2)
Choose points for each line:
- L1: y = x - 5
x = -1 → y = -6
x = 0 → y = -5
x = 1 → y = -4
x = 2 → y = -3
x = 3 → y = -2 (intersection)
x = 4 → y = -1
x = 5 → y = 0
- L2: y = 10 - 4x
x = 1 → y = 6
x = 2 → y = 2
x = 3 → y = -2 (intersection)
x = 4 → y = -6
These two lines meet at (3, -2). Here is a simple plotted grid (x from -1 to 6, y from 6 down to -6). Symbols: 1 = a point on L1, 2 = a point on L2, X = intersection (both).
x: -1 0 1 2 3 4 5 6
y= 6 . | 2 . . . . .
y= 5 . | . . . . . .
y= 4 . | . . . . . .
y= 2 . | . 2 . . . .
y= 1 . | . . . . . 1
y= 0 . 0 . . . . 1 .
y=-1 . | . . . 1 . .
y=-2 . | . . X . . .
y=-3 . | . 1 . . . .
y=-4 . | 1 . . . . .
y=-5 . 1 . . . . . .
y=-6 1 | . . . 2 . .
Legend: '|' is the y-axis (x=0), '0' marks (0,0). The X at (3,-2) is the intersection point. This confirms graphically that the two lines meet at (3, -2).
Final answer: (x, y) = (3, -2).
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