Answer:
Divide each term by
5
and simplify.
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x
=
2
y
5
+
12
5
4
x
=
28
−
3
y
Replace all occurrences of
x
in
4
x
=
28
−
3
y
with
2
y
5
+
12
5
.
x
=
2
y
5
+
12
5
4
(
2
y
5
+
12
5
)
=
28
−
3
y
Simplify
4
(
2
y
5
+
12
5
)
.
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x
=
2
y
5
+
12
5
8
y
5
+
48
5
=
28
−
3
y
Move all terms in
8
y
5
+
48
5
=
28
−
3
y
to the left side and simplify.
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x
=
2
y
5
+
12
5
23
(
y
−
4
)
5
=
0
Solve for
y
in the second equation.
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x
=
2
y
5
+
12
5
y
=
4
Replace all occurrences of
y
in
x
=
2
y
5
+
12
5
with
4
.
x
=
2
(
4
)
5
+
12
5
y
=
4
Simplify
2
(
4
)
5
+
12
5
.
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x
=
4
y
=
4
The solution to the system of equations can be represented as a point.
(
4
,
4
)
The result can be shown in multiple forms.
Point Form:
(
4
,
4
)
Equation Form:
x
=
4
,
y
=
4
Step-by-step explanation:
Answer:
1/3 or .3333333.....
Step-by-step explanation:
The answer is 4 bc you can multiply 4 with both 8 and 12. I’m sorry but I’m bad at explaining but the answer is 4 :(
Your answer is 94.
180-(28+58) gives you
180-86
=94
Answer: Adenike scored 64 marks, while Musa scored 45 marks
Step-by-step explanation: We shall start by assigning letters to each unknown variable. Let Adenike’s mark be d while Musa’s mark shall be m.
First of all, if Adenike obtained 19 marks more than Musa, then if Musa scored m, Adenike would score 19 + m (or d = 19 + m). Also if Adenike has obtained one and half her own mark (which would be 1 1/2d or 3d/2), it would have been equal to 6 times more than twice Musa’s mark (or 6 + 2m). This can be expressed as
3d/2 = 6 + 2m. So we now have a pair of simultaneous equations;
d = 19 + m ———(1)
3d/2 = 6 + 2m ———(2)
Substitute for the value of d into equation (2), if d = 19 + m
(3{19 + m})/2 = 6 + 2m
By cross multiplication we now have
3(19 + m) = 2(6 + 2m)
57 + 3m = 12 + 4m
We collect like terms and we have
57 - 12 = 4m - 3m
45 = m
We now substitute for the value of m into equation (1)
d = 19 + m
d = 19 + 45
d = 64
So Adenike scored 64 marks while Musa scored 45 marks
