49. [tex]x = \dfrac{1}{a}[/tex] and [tex]y = 3[/tex]
50. [tex]x = \dfrac{1}{2a}[/tex] and [tex]y = \dfrac{1}{b}[/tex]
Step-by-step explanation:
We are going to solve these systems of equations using the elimination method.
49. Given:
[tex]5ax + 4y = 17[/tex] (1)
[tex]\:\:ax + 7y = 22[/tex] (2)
Multiply Eqn(1) by 7 and multiply Eqn(2) by -4 to get
[tex]\:\:35ax + 28y = 119[/tex]
[tex]-4ax - 28y = -88[/tex]
Adding the two equations above, we get
[tex]31ax = 31 \Rightarrow x = \dfrac{1}{a}[/tex]
Substituting this value for a into Eqn(2), we get
[tex]y = 3[/tex]
50. Given:
[tex]4ax + \:\:by = 3[/tex] (3)
[tex]6ax + 5by = 8[/tex] (4)
Multiplying Eqn(3) by -5, we get
[tex]-20ax - 5by = -15[/tex]
[tex]\:\:\:\:\:\:6ax + 5by = \:\:8[/tex]
Adding the two equations above, we get
[tex]-14ax = -7 \Rightarrow x = \dfrac{1}{2a}[/tex]
Plugging in the value of x into Eqn(3), we get the value for y as
[tex]4a(\frac{1}{2a}) + by = 3[/tex]
or
[tex]2 + by = 3 \Rightarrow y = \dfrac{1}{b}[/tex]
