To solve the system of equations by graphing, we first rewrite each equation in slope-intercept form (y = mx + b) and then sketch the graphs.
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The first equation is already in slope-intercept form: \[ y = x + 3 \] This line has a slope of 1 and a y-intercept of 3.
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For the second equation, we start with: \[ -4x + 4y = 28 \] We can rearrange it to solve for \(y\): \[ 4y = 4x + 28 \] \[ y = x + 7 \] This line has a slope of 1 and a y-intercept of 7.
Now, we graph both equations:
Graphing:
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Graph of \(y = x + 3\):
- When \(x = 0\), \(y = 3\), so one point is (0, 3).
- When \(x = 1\), \(y = 4\), so another point is (1, 4).
- When \(x = -3\), \(y = 0\), so another point is (-3, 0).
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Graph of \(y = x + 7\):
- When \(x = 0\), \(y = 7\), so one point is (0, 7).
- When \(x = 1\), \(y = 8\), so another point is (1, 8).
- When \(x = -7\), \(y = 0\), so another point is (-7, 0).
Intersection Point:
From the graphs, we can see that the lines intersect at the point where both equations are satisfied.
To find the exact point of intersection, we set the two equations equal: \[ x + 3 = x + 7 \] This simplifies to: \[ 3 = 7 \] This is a contradiction, meaning that the lines are parallel and do not intersect. Therefore, the system of equations has no solution.
Conclusion:
The solution to the system of linear equations is: No solution.