Question 1185998: Solve for the unknown calories using the concept of linear equations
1.) 126x + 198y = 150
2.) 338x + 43y = 303 Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to solve this system of linear equations for x and y:
**Method 1: Substitution**
1. **Solve one equation for one variable:** It's easiest to solve the second equation for y:
43y = 303 - 338x
y = (303 - 338x) / 43
2. **Substitute:** Substitute this expression for y into the first equation:
126x + 198 * (303 - 338x) / 43 = 150
3. **Simplify and solve for x:**
126x + (60000 - 66924x)/43 = 150
5418x + 60000 - 66924x = 6450
-61506x = -53550
x = 53550 / 61506
x ≈ 0.872
4. **Substitute x back to find y:** Substitute the value of x into either original equation. Using the first equation is simplest.
126 * 0.872 + 198y = 150
110 + 198y = 150
198y = 40
y = 40 / 198
y ≈ 0.202
**Method 2: Elimination**
1. **Multiply equations to match coefficients:** We'll 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 => 66924x + 8514y = 60054
2. **Subtract the equations:** Subtract the first new equation from the second:
61506x = 53604
3. **Solve for x:**
x = 53604 / 61506
x ≈ 0.872
4. **Substitute x back to find y:** Substitute the value of x into either original equation.
126 * 0.872 + 198y = 150
y ≈ 0.202
**Solution:**
x ≈ 0.872
y ≈ 0.202
