Given 4x + y = 1 (equation 1) and 8x - 3y = 12 (equation 2)
Using the substitution method:
Equation 1: Subtract 4x from both sides to isolate y:
4x - 4x + y = -4x + 1
y = -4x + 1
Substitute the value of y into equation 2:
8x - 3y = 12
8x - 3(-4x + 1) = 12
8x + 12x - 3 = 12
Combine like terms:
20x - 3 = 12
Add 3 to both sides:
20x - 3 + 3 = 12 + 3
20x = 15
Divide both sides by 20 to solve for x:
20x/20 = 15/20
x = 15/20 or ¾
Substitute the value of x = ¾ into Equation 1 to solve for y:
4x + y = 1
4(¾) + y = 1
3 + y = 1
Subtract 3 from both sides to isolate y:
3 - 3 + y = 1 - 3
y = -2
Therefore, the solution of the given systems of linear equations is: x = ¾, y = -2 or (¾, -2).
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