\(5x + 3y = 9 \quad \text{...\ (I)}\)
\(2x - 3y = 12 \quad \text{...\ (II)}\)
Let’s add equations (I) and (II),
| 5x | + | 3y | = | 9 | ... (I) | |||
+ |
2x | − | 3y | = | 12 | ... (II) | ||
| \(\bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\ \)x | = | \(\bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}} \) |
\(\displaystyle \therefore x = \frac{\bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}}{\bbox[white, 5pt, border: 2px solid red]{{\color{white}{XX}}}}\)
\(\therefore x = \ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\)
Place \(x = 3\) in equation (I),
\(\therefore\ 5 \times\ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}} +\ 3y = 9\)
\(\therefore\ 3y = 9 -\ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\)
\(\therefore\ 3y = \ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\)
\(\displaystyle \therefore y = \frac{\bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}}{3}\)
\(\therefore\ y = \ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\)
∴ Solution is \((x, y) = (\ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\:,\ \bbox[white, 5pt, border: 2px solid red]{\color{white}{XX}}\ )\)
The completed activity is given below:
\(5x + 3y = 9 \quad \text{...\ (I)}\)
\(2x - 3y = 12 \quad \text{...\ (II)}\)
Let’s add equations (I) and (II),
| 5x | + | 3y | = | 9 | ... (I) | ||
+ |
2x | − | 3y | = | 12 | ... (II) | |
| \(\bbox[white, 5pt, border: 2px solid red]{7}\ \)x | = | \(\bbox[white, 5pt, border: 2px solid red]{21} \) |
\(\displaystyle \therefore x = \frac{\bbox[white, 5pt, border: 2px solid red]{21}}{\bbox[white, 5pt, border: 2px solid red]{7}}\)
\(\therefore x = \ \bbox[white, 5pt, border: 2px solid red]{3}\)
Place \(x = 3\) in equation (I),
\(\therefore\ 5 \times\ \bbox[white, 5pt, border: 2px solid red]{3} +\ 3y = 9\)
\(\therefore\ 15 + 3y = 9\)
\(\therefore\ 3y = 9 -\ \bbox[white, 5pt, border: 2px solid red]{15}\)
\(\therefore\ 3y = \ \bbox[white, 5pt, border: 2px solid red]{-\:6}\)
\(\displaystyle \therefore y = \frac{\bbox[white, 5pt, border: 2px solid red]{-\:6}}{3}\)
\(\therefore\ y = \ \bbox[white, 5pt, border: 2px solid red]{-\:2}\)
∴ Solution is \((x, y) = (\ \bbox[white, 5pt, border: 2px solid red]{3}\:,\ \bbox[white, 5pt, border: 2px solid red]{-\:2}\ )\)
\(3a + 5b = 26\) ... (i)
and \(a + 5b = 22\) ... (ii)
Subtracting (ii) from (i),
| 3a | + | 5b | = | 26 | ... (i) | |||
− |
⊕ | a | ⊕ | 5b | = | ⊕ | 22 | ... (ii) |
| − | − | − | ||||||
| 2a | = | 4 |
∴ \(\displaystyle a = \frac{\cancelto{2}{4}}{\cancelto{1}{2}}\)
∴ \(a = 2\) ... (iii)
Substituting the value of \(a\) in (ii),
\(a + 5b = 22\) ... (ii)
∴ \(2 + 5b = 22\)
∴ \(5b = 22 - 2\)
∴ \(5b = 20\)
∴ \(\displaystyle b = \frac{\cancelto{4}{20}}{\cancelto{1}{5}}\)
∴ \(b = 4\) ... (iv)
∴ (2, 4) is the solution of the given simultaneous equations.
\(x + 7y = 10\)
∴ \(x = 10 - 7y\) ... (i)
and \(3x - 2y = 7\) ... (ii)
Substituting the value of \(x\) in (ii),
\(3x - 2y = 7\) ... (ii)
∴ \(3(10 - 7y) - 2y = 7\)
∴ \(30 - 21y - 2y = 7\)
∴ \(30 - 23y = 7\)
∴ \(-\:23y = 7 - 30\)
∴ \(-\:23y = -\:23\)
∴ \(23y = 23\)
∴ \(\displaystyle y = \frac{\cancelto{1}{23}}{\cancelto{1}{23}}\)
∴ \(\displaystyle y = 1\) ... (iii)
Substituting the value of \(y\) in (i),
\(x = 10 - 7y\) ... (i)
∴ \(x = 10 - 7 \times 1\)
∴ \(x = 10 - 7\)
∴ \(x = 3\) ... (iv)
∴ (3, 1) is the solution of the given simultaneous equations.
\(2x - 3y = 9\) ... (i)
and \(2x + y = 13\) ... (ii)
Subtracting (ii) from (i),
| 2x | − | 3y | = | 9 | ... (i) | |||
− |
⊕ | 2x | ⊕ | y | = | ⊕ | 13 | ... (ii) |
| − | − | − | ||||||
| − | 4y | = | − | 4 |
i.e. \(4y = 4\)
∴ \(\displaystyle y = \frac{\cancelto{1}{4}}{\cancelto{1}{4}}\)
∴ \(y = 1\) ... (iii)
Substituting the value of \(y\) in (ii),
\(2x + y = 13\) ... (ii)
∴ \(2x + 1 = 13\)
∴ \(2x = 13 - 1\)
∴ \(2x = 12\)
∴ \(\displaystyle x = \frac{\cancelto{6}{12}}{\cancelto{1}{2}}\)
∴ \(x = 6\) ... (iv)
∴ (6, 1) is the solution of the given simultaneous equations.
\(5m - 3n = 19\) ... (i)
and \(m - 6n = -\:7\) ... (ii)
Multiplying (i) by 2,
\(10m - 6n = 38\) ... (iii)
Subtracting (ii) from (iii),
| 10m | − | 6n | = | 38 | ... (iii) | |||
− |
⊕ | m | ⊖ | 6n | = | ⊖ | 7 | ... (ii) |
| − | + | + | ||||||
| 9m | = | 45 |
∴ \(\displaystyle m = \frac{\cancelto{5}{45}}{\cancelto{1}{9}}\)
∴ \(m = 5\) ... (iv)
Substituting the value of \(m\) in (ii),
\(m - 6n = -\:7\) ... (ii)
∴ \(5 - 6n = -\:7\)
∴ \(-\:6n = -\:7\:-\:5\)
∴ \(-\:6n = -\:12\)
i.e. \(6n = 12\)
∴ \(\displaystyle n = \frac{\cancelto{2}{12}}{\cancelto{1}{6}}\)
∴ \(n = 2\) ... (v)
∴ (5, 2) is the solution of the given simultaneous equations.
\(5x + 2y =\:-\:3\) ... (i)
and \(x + 5y = 4\)
∴ \(x = 4\:-\:5y\) ... (ii)
Substituting the value of \(x\) in (i),
\(5x + 2y =\:-\:3\) ... (i)
∴ \(5(4\:-\:5y) + 2y =\:-\:3\)
∴ \(20\:-\:25y + 2y =\:-\:3\)
∴ \(20\:-\:23y =\:-\:3\)
∴ \(-\:23y =\:-\:3\:-\:20\)
∴ \(-\:23y =\:-\:23\)
i.e. \(23y = 23\)
∴ \(\displaystyle y = \frac{\cancelto{1}{23}}{\cancelto{1}{23}}\)
∴ \(y = 1\) ... (iii)
Substituting the value of \(y\) in (ii),
\(x = 4\:-\:5y\) ... (ii)
∴ \(x = 4\:-\:5 \times 1\)
∴ \(x = 4\:-\:5\)
∴ \(x =\:-\:1\) ... (iv)
∴ ( − 1, 1) is the solution of the given simultaneous equations.
\(\displaystyle \frac{1}{3}x + y = \frac{10}{3}\)
Multiplying both sides by 3,
\(\displaystyle \frac{1}{3}x \times {3} + y \times {3} = \frac{10}{3} \times {3}\)
∴ \(\displaystyle \frac{1}{\cancel{3}}x \times {\cancel{3}} + y \times {3} = \frac{10}{\cancel{3}} \times {\cancel{3}}\)
∴ \(x + 3y = 10\)
i.e. \(x = 10 - 3y\) ... (i)
and \(\displaystyle 2x + \frac{1}{4}y = \frac{11}{4}\)
Multiplying both sides by 4,
\(\displaystyle 2x \times {4} + \frac{1}{\cancel{4}}y \times {\cancel{4}} = \frac{11}{\cancel{4}} \times {\cancel{4}}\)
∴ \(\displaystyle 8x + y = 11\) ... (ii)
Substituting the value of \(x\) in (ii),
\(8x + y = 11\) ... (ii)
∴ \(8(10\:-\:3y) + y = 11\)
∴ \(80\:-\:24y + y = 11\)
∴ \(80\:-\:23y = 11\)
∴ \(\:-\:23y = 11\:-\:80\)
∴ \(\:-\:23y = \:-\:69\)
i.e. \(23y = 69\)
∴ \(\displaystyle y = \frac{\cancelto{3}{69}}{\cancelto{1}{23}}\)
∴ \(y = 3\) ... (iii)
Substituting the value of \(y\) in (i),
\(x = 10\:-\:3y\) ... (ii)
∴ \(x = 10\:-\:3 \times 3\)
∴ \(x = 10\:-\:9\)
∴ \(x = 1\) ... (iv)
∴ (1, 3) is the solution of the given simultaneous equations.
\(99x + 101y = 499\) ... (i)
And, \(101x + 99y = 501\) ... (ii)
Adding (i) and (ii),
| 99x | + | 101y | = | 499 | ... (i) | |||
+ |
101x | + | 99y | = | 501 | ... (ii) | ||
| 200x | + | 200y | = | 1000 |
Dividing both sides by 200,
\(x + y = 5\) ... (iii)
Subtracting (i) from (ii),
| 101x | + | 99y | = | 501 | ... (ii) | |||
− |
⊕ | 99x | ⊕ | 101y | = | ⊕ | 499 | ... (i) |
| − | − | − | ||||||
| 2x | − | 2y | = | 2 |
Dividing both sides by 2,
\(x -\: y = 1\) ... (iv)
Adding (iii) and (iv),
| x | + | y | = | 5 | ... (iii) | |||
+ |
x | − | y | = | 1 | ... (iv) | ||
| 2x | = | 6 |
∴ \(\displaystyle x = \frac{\cancelto{3}{6}}{\cancelto{1}{2}}\)
∴ \(x = 3\) ... (v)
Substituting the value of \(x\) in (iii),
\(x + y = 5\) ... (iii)
∴ \(3 + y = 5\)
∴ \(y = 5\:-\:3\)
∴ \(y = 2\) ... (vi)
∴ (3, 2) is the solution of the given equations.
\(49x - 57y = 172\) ... (i)
And, \(57x - 49y = 252\) ... (ii)
Adding (i) and (ii),
| 49x | − | 57y | = | 172 | ... (i) | |||
+ |
57x | − | 49y | = | 252 | ... (ii) | ||
| 106x | − | 106y | = | 424 |
Dividing both sides by 106,
\(x - y = 4\) ... (iii)
Subtracting (i) from (ii),
| 57x | − | 49y | = | 252 | ... (ii) | |||
− |
⊕ | 49x | ⊖ | 57y | = | ⊕ | 172 | ... (i) |
| − | + | − | ||||||
| 8x | + | 8y | = | 80 |
Dividing both sides by 8,
\(x + y = 10\) ... (iv)
Adding (iii) and (iv),
| x | − | y | = | 4 | ... (iii) | |||
+ |
x | + | y | = | 10 | ... (iv) | ||
| 2x | = | 14 |
∴ \(\displaystyle x = \frac{\cancelto{7}{14}}{\cancelto{1}{2}}\)
∴ \(x = 7\) ... (v)
Substituting the value of \(x\) in (iv),
\(x + y = 10\) ... (iv)
∴ \(7 + y = 10\)
∴ \(y = 10\:-\:7\)
∴ \(y = 3\) ... (vi)
∴ (7, 3) is the solution of the given equations.
This page was last modified on
12 May 2026 at 15:00