To find the equation of the line passing through the intersection of [tex]\(3x + y = 7\)[/tex] and [tex]\(3y = 4x - 5\)[/tex], and parallel to [tex]\(2x - y = 3\)[/tex], follow these steps:
1. Find the intersection point of [tex]\(3x + y = 7\)[/tex] and [tex]\(3y = 4x - 5\)[/tex]:
- From the equation [tex]\(3x + y = 7\)[/tex]:
[tex]\[ y = 7 - 3x \][/tex]
- Substitute [tex]\(y = 7 - 3x\)[/tex] into the equation [tex]\(3y = 4x - 5\)[/tex]:
[tex]\[ 3(7 - 3x) = 4x - 5 \][/tex]
[tex]\[ 21 - 9x = 4x - 5 \][/tex]
[tex]\[ 21 + 5 = 4x + 9x \][/tex]
[tex]\[ 26 = 13x \][/tex]
[tex]\[ x = 2 \][/tex]
- Substitute [tex]\(x = 2\)[/tex] back into [tex]\(y = 7 - 3x\)[/tex]:
[tex]\[ y = 7 - 3(2) \][/tex]
[tex]\[ y = 1 \][/tex]
- Hence, the intersection point is [tex]\((2, 1)\)[/tex].
2. Determine the equation of the line parallel to [tex]\(2x - y = 3\)[/tex] and passing through [tex]\((2, 1)\)[/tex]:
- The general form of the equation for a line parallel to [tex]\(2x - y = 3\)[/tex] has the same slope, thus:
[tex]\[ 2x - y = C \][/tex]
- We need to find the constant [tex]\(C\)[/tex] using the point [tex]\((2, 1)\)[/tex]:
[tex]\[ 2(2) - 1 = C \][/tex]
[tex]\[ 4 - 1 = C \][/tex]
[tex]\[ C = 3 \][/tex]
- Therefore, the equation of the line is:
[tex]\[ 2x - y = 3 \][/tex]
Since this line is already in the form [tex]\(2x - y = 3\)[/tex], the required equation is:
[tex]\[ 2x - y - 3 = 0 \][/tex]
