4х+у=10 х+3у=-3

To solve the system of equations: 4𝑥+𝑦=10 𝑥+3𝑦=−3

We can use the method of substitution or elimination.

Method 1: Substitution First, let's solve one of the equations for 𝑥 or 𝑦 and substitute it into the other equation.

From the second equation, we can solve for 𝑥: 𝑥 = -3 - 3𝑦

Substituting this value of 𝑥 into the first equation: 4(-3 - 3𝑦) + 𝑦 = 10 -12 - 12𝑦 + 𝑦 = 10 -12𝑦 + 𝑦 = 10 + 12 -11𝑦 = 22 𝑦 = -2

Now, substitute 𝑦 = -2 back into the second equation to find 𝑥: 𝑥 + 3(-2) = -3 𝑥 - 6 = -3 𝑥 = 3

So the solution to the system of equations is 𝑥 = 3 and 𝑦 = -2.

Method 2: Elimination Multiply the second equation by 4 to create a system where the coefficients of 𝑥 are the same:

4(𝑥 + 3𝑦) = 4(-3) 4𝑥 + 12𝑦 = -12

Now we have two equations: 4𝑥 + 𝑦 = 10 4𝑥 + 12𝑦 = -12

Subtract the first equation from the second equation to eliminate 𝑥: (4𝑥 + 12𝑦) - (4𝑥 + 𝑦) = -12 - 10 11𝑦 = -22 𝑦 = -2

Now, substitute 𝑦 = -2 back into the first equation to find 𝑥: 4𝑥 + (-2) = 10 4𝑥 - 2 = 10 4𝑥 = 12 𝑥 = 3

So the solution to the system of equations is 𝑥 = 3 and 𝑦 = -2.

Both methods yield the same solution: 𝑥 = 3 and 𝑦 = -2.