solve for x and y using substitution method 2x-3y/4=0 and 2x +y-8 =0
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Answered by
2
Answer: x=12; y=32/7
Step-by-step explanation: taking 1,
2x=3y/4
= x= 3y/8...i
substituting this in 2
= 2(3y/8) + y -8=0
= 3y/4 +y -8 =0
=7y/4=8
= 7y=32
=y=32/7
substituting y in (i)
=x = 3*32/8
=x= 12
ArtemisW:
hey..sorry.. x is wrong.
should be 12/7
sorry..it was a silly mistake
Answered by
1
2x×4/4 - 3y/4 =0
8x - 3y = 0 equation 1
2x + y = 8 equation 2
eq 2 - eq1
2x + y = 8 ×4
8x + 4y = 32
-8x+3y = 0
___________
7y = 32
y=32/7
put the value of y in eq1
8x - 3×32/7 = 0
8x = 3× 32/7
x = 3× 32/7×8
x= 3×4/7
x= 12/7
8x - 3y = 0 equation 1
2x + y = 8 equation 2
eq 2 - eq1
2x + y = 8 ×4
8x + 4y = 32
-8x+3y = 0
___________
7y = 32
y=32/7
put the value of y in eq1
8x - 3×32/7 = 0
8x = 3× 32/7
x = 3× 32/7×8
x= 3×4/7
x= 12/7
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