Question

solve the system of inequalities by graphing. if there
x + y < 2
2x + y < 4

Explanation:

Step1: Rewrite inequalities as lines

For \(x + y < 2\), rewrite as \(y < -x + 2\). The boundary line is \(y = -x + 2\) (dashed, since \(<\)), with slope \(-1\), y - intercept \(2\), x - intercept \(2\).
For \(2x + y < 4\), rewrite as \(y < -2x + 4\). The boundary line is \(y = -2x + 4\) (dashed), slope \(-2\), y - intercept \(4\), x - intercept \(2\).

Step2: Graph the boundary lines

Plot \(y = -x + 2\): passes through \((0, 2)\) and \((2, 0)\), draw dashed line.
Plot \(y = -2x + 4\): passes through \((0, 4)\) and \((2, 0)\), draw dashed line.

Step3: Determine shading regions

For \(y < -x + 2\), shade below the line (test \((0, 0)\): \(0 < 2\), true).
For \(y < -2x + 4\), shade below the line (test \((0, 0)\): \(0 < 4\), true).

Step4: Find the intersection region

The solution is the overlapping shaded region (below both lines, bounded by the two dashed lines and their intersection at \((2, 0)\)).

Answer:

The solution is the region below both dashed lines \(y = -x + 2\) and \(y = -2x + 4\), including all points \((x, y)\) that satisfy both \(x + y < 2\) and \(2x + y < 4\) (graphically, the area where the two shaded regions overlap, bounded by the two lines and their intersection at \((2, 0)\)).