divide :
(y^2 -7y +12 ) (y+4) ÷( y-4)
Answers
Answer:
The values of x and y in the given equation by substitution method is x=20 and y=30 for x\neq 0x
=0 and y\neq 0y
=0
Step-by-step explanation:
Given equations are \frac{4}{x}+\frac{9}{y}=\frac{1}{2}\hfill (1)
x
4
+
y
9
=
2
1
\hfill(1)
and \frac{12}{x}+\frac{12}{y}=1\hfill (2)
x
12
+
y
12
=1\hfill(2) for x\neq 0x
=0 and y\neq 0y
=0
Now solving the given equations by substitution method :
Let a=\frac{1}{x}a=
x
1
and b=\frac{1}{y}b=
y
1
for x\neq 0x
=0 and y\neq 0y
=0
Equation (1) implies 4a+9b=\frac{1}{2}\hfill (3)4a+9b=
2
1
\hfill(3)
Equation (2) implies 12a+12b=1\hfill (4)12a+12b=1\hfill(4)
From equation (3) 4a+9b=\frac{1}{2}4a+9b=
2
1
4a=\frac{1}{2}-9b4a=
2
1
−9b
a=\frac{\frac{1}{2}-9b}{4}a=
4
2
1
−9b
a=\frac{1}{8}-\frac{9b}{4}a=
8
1
−
4
9b
Now substitute the value of "a" in equation (4)
12(\frac{1}{8}-\frac{9b}{4})+12b=112(
8
1
−
4
9b
)+12b=1
\frac{12}{8}-\frac{108b}{4}+12b=1
8
12
−
4
108b
+12b=1
\frac{3}{2}-27b+12b=1
2
3
−27b+12b=1
-15b=1-\frac{3}{2}−15b=1−
2
3
-15b=\frac{2-3}{2}−15b=
2
2−3
-15b=-\frac{1}{2}−15b=−
2
1
b=\frac{1}{2\times 15}b=
2×15
1
b=\frac{1}{30}b=
30
1
Since b=\frac{1}{y}b=
y
1
for y\neq 0y
=0
b=\frac{1}{30}=\frac{1}{y}b=
30
1
=
y
1
for y\neq 0y
=0
Therefore y=30
Substitute the value of b in equation a=\frac{1}{8}-\frac{9b}{4}a=
8
1
−
4
9b
we get
a=\frac{1}{8}-\frac{9(\frac{1}{30})}{4}a=
8
1
−
4
9(
30
1
)
=\frac{1}{8}-\frac{\frac{9}{30}}{4}=
8
1
−
4
30
9
=\frac{1}{8}-\frac{9}{120}=
8
1
−
120
9
=\frac{1}{8}-\frac{3}{40}=
8
1
−
40
3
=\frac{5-3}{40}=
40
5−3
=\frac{2}{40}=
40
2
a=\frac{1}{20}a=
20
1
Since a=\frac{1}{x}a=
x
1
for x\neq 0x
=0
a=\frac{1}{20}=\frac{1}{x}a=
20
1
=
x
1
for x\neq 0x
=0
Therefore x=20
Therefore x=20 and y=30
Answer:
Step-by-step explanation:
