Which of the following is not a postulate or theorem used to prove two triangles are congruent
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To solve inequalities algebraically, first, you have to graph the lines disregarding the inequalities for a while.
For line 3x-y-7 <0, let's disregard < first and change it to =, such that 3x-y-7=0. Rearranging, y=3x-7. If you plot this equation, that would be the blue line in the equation. To know which of side of the line is the solution, you substitute a random point into the equality. For example, let's use point (5,20).
3x-y-7<0
3(5)-20-7<0
-12<0, this is true. Therefore, all the space to the left of the blue line is a solution (blue region).
We do the same for the other equation (orange line). Let's use the same point (5,20) to test.
x+5y+3≥0
5+5(20)+3≥0
108≥0, this is true, Therefore, everything above the orange line is a solution (orange region).
The overlapped area of the two shaded regions is presented as green in the picture. This is the exact solution of this system of linear equations.
For line 3x-y-7 <0, let's disregard < first and change it to =, such that 3x-y-7=0. Rearranging, y=3x-7. If you plot this equation, that would be the blue line in the equation. To know which of side of the line is the solution, you substitute a random point into the equality. For example, let's use point (5,20).
3x-y-7<0
3(5)-20-7<0
-12<0, this is true. Therefore, all the space to the left of the blue line is a solution (blue region).
We do the same for the other equation (orange line). Let's use the same point (5,20) to test.
x+5y+3≥0
5+5(20)+3≥0
108≥0, this is true, Therefore, everything above the orange line is a solution (orange region).
The overlapped area of the two shaded regions is presented as green in the picture. This is the exact solution of this system of linear equations.