Answer
The system of equations are as follows;
y = 9 + 3z......(1)
y = 51 - 3z.....(2)
Subtracting equation(2) from equation(1), we get,
y - y = (9 + 3z) - (51 - 3z)
Or, 0 = 9 + 3z - 51 + 3z
Or, 0 = 6z - 42
Or, 6z - 42 = 0
Or, 6z = 42
Or, z = 7
Now, putting the value of 'z' in equation(1), we get,
y = 9 + 3 * 7
Or, y = 9 + 21
Or, y = 30
So, Value of z = 7 ; Value of y = 30
System of Equations | Value of z | Value of y |
y = 9 + 3z y = 51 - 3z |
7 | 30 |
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The system of equations are as follows;
9a + 3b = 30.......(1)
8a + 4b = 28.......(2)
Multiplying equation(1) by the co-efficient of 'a' of equation(2), i.e. '8', we get,
(9a + 3b = 30) * 8
Or, 72a + 24b = 240.....(1')
Multiplying equation(2) by the co-efficient of 'a' of equation(1), i.e. '9', we get,
72a + 36b = 252.......(2')
Now, subtracting equation(2') from equation(1'), we get,
(72a + 24b) - (72a + 36b) = 240 - 252
Or, 72a + 24b - 72a - 36b = -12
Or, 24b - 36b = -12
Or, -12b = -12
Or, b = -12 / - 12
Or, b = 1
Now, putting the value of 'b' in equation(1), we get,
9a + 3 * 1 = 30
Or, 9a + 3 = 30
Or, 9a = 30 - 3
Or, 9a = 27
Or, a = 27 / 9
Or, a = 3
So, Value of a = 3 ; Value of b = 1
System of Equations | Value of a | Value of b |
9a + 3b = 30 8a + 4b = 28 |
3 | 1 |
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The system of equations are as follows;
p = 6 - 2q......(1)
p = 2 + q.....(2)
Now, let us build two tables for above two equations;
Table-1 for equation, p = 6 - 2q
q | p |
0 | 6 |
1 | 4 |
2 | 2 |
3 | 0 |
Table-2 for equation,.p = 2 + q
q | p |
0 | 2 |
1 | 3 |
2 | 4 |
3 | 5 |
4 | 6 |
5 | 7 |
6 | 8 |
7 | 9 |
8 | 10 |
11 | 13 |
12 | 14 |
Based on the above tables, we get the following graph;
The values of p and q that solve these two equations simultaneously can be seen on the graph as p=3.3(approx), q = 1.3(approx.).
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