We are given the linear systems of equations given below:
3x -y = -6 color(blue)(Eqn.1)
4x +3y = 29 color(blue)(Eqn.2)
Multiply color(blue)(Eqn.1) by 3
Hence, color(blue)(Eqn.1) yields color(blue)(Eqn.3)
9x -3y = -18 color(blue)(Eqn.3)
4x +3y = 29 color(blue)(Eqn.2)
When we add color(blue)(Eqn.3) and color(blue)(Eqn.2) we get
9x -cancel(3y) = -18 color(blue)(Eqn.3)
4x +cancel(3y) = 29 color(blue)(Eqn.2)
rArr 13x = 11
Therefore, color(red)(x = 11/13)
Substitute this value of color(red)(x) in color(blue)(Eqn.1)
3x -y = -6 color(blue)(Eqn.1)
rArr 3(11/13) - y = -6
rArr (33/13) - y = -6
rArr - y = -6 - (33/13)
rArr - y = (-78 - 33)/13
rArr - y = -111/13
Divide both sides by -1 to get
rArr y =111/13
Hence, our final solutions are : -
color(red)[(x = 11/13) and (y = 111/13) OR
We can simplify the fractions and write the solutions as
color(red)[(x ~~ 0.8) and (y ~~ 8.5)
