Final answer:
To solve the simultaneous equations, equation (2) is multiplied by 3 to match the y-coefficient with equation (1). After subtracting the updated equation (2) from equation (1), we find that x = 0.5. Substituting this value into any original equation gives y = 3, resulting in the solution x = 0.5 and y = 3.
Step-by-step explanation:
To solve the given simultaneous linear equations by multiplying equation (2) by a constant, we apply the method of elimination. The two equations are: (1) 3y + 8x = 13 and (2) y + 6x = 6. The goal is to achieve a common coefficient for either x or y in both equations, which enables us to subtract one equation from the other and solve for the remaining variable.
First, we identify a suitable constant to multiply the second equation by, in order to have the same coefficients for either x or y. Multiplying the second equation by 3 (the coefficient of y in the first equation) gives us: 3*(y + 6x) = 3*6, which simplifies to 3y + 18x = 18. We now have two equations with the same coefficient for y:
- 3y + 8x = 13 (Equation 1)
- 3y + 18x = 18 (Equation 2 multiplied by 3)
Next, we subtract Equation 1 from the modified Equation 2:
(3y + 18x) - (3y + 8x) = 18 - 13
This gives us 10x = 5. Solving for x, we get:
x = 5/10
x = 1/2 or 0.5
Now that we have the value of x, we can substitute it back into either of the original equations to find the value of y. Using Equation (2):
y + 6*(1/2) = 6
y + 3 = 6
y = 6 - 3
y = 3
Thus, the solution to the simultaneous equations is x = 0.5 and y = 3.
In the context of linear equations similar to the ones provided in the problem, such equations can represent various real-world scenarios, such as the relationship between cost and attendance, hours required and square footage, and so on, where one variable depends on the other.