Absolutely, let's solve this step-by-step.
Step 1: Solve the equations for [tex]\( x \)[/tex]
We have two linear equations:
1. [tex]\( \Delta = 3x - 2 \)[/tex]
2. [tex]\( \Delta = 6x + 7 \)[/tex]
Step 2: From the first equation, solve for [tex]\( x \)[/tex]:
[tex]\[ 3x - 2 = 0 \][/tex]
[tex]\[ 3x = 2 \][/tex]
[tex]\[ x = \frac{2}{3} \][/tex]
Step 3: From the second equation, solve for [tex]\( x \)[/tex]:
[tex]\[ 6x + 7 = 0 \][/tex]
[tex]\[ 6x = -7 \][/tex]
[tex]\[ x = -\frac{7}{6} \][/tex]
So, the solutions to the equations are [tex]\( x = \frac{2}{3} \)[/tex] and [tex]\( x = -\frac{7}{6} \)[/tex].
Step 4: Calculate the values for each case using the solution [tex]\( x = \frac{2}{3} \)[/tex]:
Since the function given is [tex]\(\Delta = 3x - 2\)[/tex], we calculate it for [tex]\( x = \frac{2}{3} \)[/tex]:
[tex]\[ \Delta = 3\left(\frac{2}{3}\right) - 2 \][/tex]
[tex]\[ \Delta = 2 - 2 \][/tex]
[tex]\[ \Delta = 0 \][/tex]
Now, for each A, B, C, D, and E, we calculate the absolute value difference from 28, 29, 30, 31, and 32 respectively:
A) [tex]\[ \left|28 - 0\right| = 28 \][/tex]
B) [tex]\[ \left|29 - 0\right| = 29 \][/tex]
C) [tex]\[ \left|30 - 0\right| = 30 \][/tex]
D) [tex]\[ \left|31 - 0\right| = 31 \][/tex]
E) [tex]\[ \left|32 - 0\right| = 32 \][/tex]
Hence, the values are:
[tex]\[ A = 28 \][/tex]
[tex]\[ B = 29 \][/tex]
[tex]\[ C = 30 \][/tex]
[tex]\[ D = 31 \][/tex]
[tex]\[ E = 32 \][/tex]
In summary, the solution gives us the calculated values of 28, 29, 30, 31, and 32.
