SOLUTION: Solve for the unknown calories using the concept of linear equations 1.) 126x + 198y = 150 2.) 338x + 43y = 303 With graphing method and subtitution, elimination, determinants

Question 1186000: Solve for the unknown calories using the concept of linear equations
1.) 126x + 198y = 150
2.) 338x + 43y = 303
With graphing method and subtitution, elimination, determinants
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to solve the system of equations using the methods you requested:
**1. Graphing Method:**
1. **Rewrite the equations in slope-intercept form (y = mx + b):**
* 126x + 198y = 150 => y = (-126/198)x + (150/198) => y ≈ -0.636x + 0.758
* 338x + 43y = 303 => y = (-338/43)x + (303/43) => y ≈ -7.86x + 7.047
2. **Plot the lines:** Graph both equations on the same coordinate plane. The point where the lines intersect is the solution.
3. **Approximate the solution:** From the graph, you'll see that the lines intersect somewhere around x ≈ 0.8 and y ≈ 0.3. Graphing isn't always perfectly precise, so these are just estimates.
**2. Substitution Method:**
1. **Solve one equation for one variable:** Let's solve the first equation for y:
y = (-126/198)x + (150/198)
2. **Substitute:** Substitute this expression for y into the second equation:
338x + 43((-126/198)x + (150/198)) = 303
3. **Simplify and solve for x:**
338x - (5418/198)x + (6450/198) = 303
(66864/198)x - (5418/198)x = (60054/198) - (6450/198)
(61446/198)x = (53604/198)
x = 53604 / 61446
x ≈ 0.872
4. **Substitute x back to find y:** Substitute the value of x into either of the original equations. Using the first equation is simpler.
126(0.872) + 198y = 150
110 + 198y = 150
198y = 40
y ≈ 0.202
**3. Elimination Method:**
1. **Multiply equations to match coefficients:** We want to eliminate either x or y. Let's eliminate y. Multiply the first equation by 43 and the second equation by 198:
* (126x + 198y = 150) * 43 => 5418x + 8514y = 6450
* (338x + 43y = 303) * 198 => 66864x + 8514y = 60054
2. **Subtract the equations:** Subtract the first new equation from the second:
61446x = 53604
3. **Solve for x:**
x ≈ 0.872
4. **Substitute x back to find y:** Substitute the value of x into either of the original equations.
126(0.872) + 198y = 150
y ≈ 0.202
**4. Determinant Method (Cramer's Rule):**
1. **Set up the coefficient matrix (D) and the matrices for x (Dx) and y (Dy):**
D = | 126 198 |
| 338 43 |
Dx = | 150 198 |
| 303 43 |
Dy = | 126 150 |
| 338 303 |
2. **Calculate the determinants:**
* det(D) = (126 * 43) - (198 * 338) = -61446
* det(Dx) = (150 * 43) - (198 * 303) = -53604
* det(Dy) = (126 * 303) - (150 * 338) = -12300
3. **Solve for x and y:**
x = det(Dx) / det(D) = -53604 / -61446 ≈ 0.872
y = det(Dy) / det(D) = -12300 / -61446 ≈ 0.200
**Solution:**
The solution using all methods converges to approximately:
* x ≈ 0.872
* y ≈ 0.202

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