MathLabs

Problem 5

Solve the following system of equations for real numbers x1x_1, x2x_2, x3x_3, x4x_4: ∣a1−a2∣x2+∣a1−a3∣x3+∣a1−a4∣x4=1∣a2−a1∣x1+∣a2−a3∣x3+∣a2−a4∣x4=1∣a3−a1∣x1+∣a3−a2∣x2+∣a3−a4∣x4=1∣a4−a1∣x1+∣a4−a2∣x2+∣a4−a3∣x3=1\begin{aligned} |a_1-a_2|x_2 + |a_1-a_3|x_3 + |a_1-a_4|x_4 &= 1 \\ |a_2-a_1|x_1 + |a_2-a_3|x_3 + |a_2-a_4|x_4 &= 1 \\ |a_3-a_1|x_1 + |a_3-a_2|x_2 + |a_3-a_4|x_4 &= 1 \\ |a_4-a_1|x_1 + |a_4-a_2|x_2 + |a_4-a_3|x_3 &= 1 \end{aligned} where a1a_1, a2a_2, a3a_3, a4a_4 are mutually distinct real numbers.
Step 2 of 7: Subtract equation 1 minus equation 2
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In plain words

Subtracting two equations that share most of their coefficients cancels the parts that depend on the actual values of the aia_i, leaving a pure relation among the xix_i.

−x1+x2+x3+x4=0-x_1+x_2+x_3+x_4=0
Detailed analysis

With absolute values resolved, equation 1 is (a1−a2)x2+(a1−a3)x3+(a1−a4)x4=1(a_1-a_2)x_2+(a_1-a_3)x_3+(a_1-a_4)x_4=1 and equation 2 is (a1−a2)x1+(a2−a3)x3+(a2−a4)x4=1(a_1-a_2)x_1+(a_2-a_3)x_3+(a_2-a_4)x_4=1. Subtracting and dividing by a1−a2≠0a_1-a_2\ne0 leaves −x1+x2+x3+x4=0-x_1+x_2+x_3+x_4=0.