Solve the system of equations. {:[-5y+6x=40],[3y-8x=-46],[x=◻],[y=◻]:}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can eliminate 'y' by multiplying the first equation by 3 and the second equation by 5 to get the coefficients of 'y' to be opposites.Multiply first equation by 3: Multiply the first equation by 3: 3(-5y + 6x) = 3(40), which gives us -15y + 18x = 120.Multiply second equation by 5: Multiply the second equation by 5: 5(3y - 8x) = 5(-46), which gives us 15y - 40x = -230.Add equations to eliminate 'y': Add the new equations from Step 2 and Step 3 to eliminate 'y': (-15y + 18x) + (15y - 40x) = 120 + (-230).Perform addition: Perform the addition: -15y + 15y + 18x - 40x = 120 - 230, which simplifies to -22x = -110.Solve for 'x': Solve for 'x'. Dividing both sides of the equation by -22 gives us x = -110 / -22, which simplifies to x = 5.Substitute x = 5: Substitute x = 5 into one of the original equations to solve for 'y'. We can use the first equation: -5y + 6(5) = 40.Substitute x into equation: Substitute x into the equation: -5y + 30 = 40. Subtract 30 from both sides to get -5y = 10.Solve for 'y': Solve for 'y'. Dividing both sides of the equation by -5 gives us y = 10 / -5, which simplifies to y = -2.Write the solution: Write the solution as a coordinate point. The solution is (5, -2).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the system of equations.

{:[-5y+6x=40],[3y-8x=-46],[x=◻],[y=◻]:}

Solve the system of equations. $\newline$ $\begin{array}{l} -5 y+6 x=40 \\ 3 y-8 x=-46 \\ x=\square \\ y=\square \end{array}$

View step-by-step help

Home

Math Problems

Algebra 2

Solve a system of equations using elimination

Full solution

Q. Solve the system of equations.

\newline

\begin{array}{l} -5 y+6 x=40 \\ 3 y-8 x=-46 \\ x=\square \\ y=\square \end{array}

Identify variable to eliminate: Identify the variable to eliminate. In this case, we can eliminate ' extit{y}' by multiplying the first equation by $3$ and the second equation by $5$ to get the coefficients of ' extit{y}' to be opposites.
Multiply first equation by $3$ : Multiply the first equation by $3$ : $3(-5y + 6x) = 3(40)$ , which gives us $-15y + 18x = 120$ .
Multiply second equation by $5$ : Multiply the second equation by $5$ : $5(3y - 8x) = 5(-46)$ , which gives us $15y - 40x = -230$ .
Add equations to eliminate 'y': Add the new equations from Step $2$ and Step $3$ to eliminate 'y': $(-15y + 18x) + (15y - 40x) = 120 + (-230)$ .
Perform addition: Perform the addition: $-15y + 15y + 18x - 40x = 120 - 230$ , which simplifies to $-22x = -110$ .
Solve for 'x': Solve for 'x'. Dividing both sides of the equation by $-22$ gives us $x = \frac{-110}{-22}$ , which simplifies to $x = 5$ .
Substitute $x = 5$ : Substitute $x = 5$ into one of the original equations to solve for ' $y$ '. We can use the first equation: $-5y + 6(5) = 40$ .
Substitute $x$ into equation: Substitute $x$ into the equation: $-5y + 30 = 40$ . Subtract $30$ from both sides to get $-5y = 10$ .
Solve for 'y': Solve for 'y'. Dividing both sides of the equation by $-5$ gives us $y = \frac{10}{-5}$ , which simplifies to $y = -2$ .
Write the solution: Write the solution as a coordinate point. The solution is $(5, -2)$ .