How do you solve #5x - 1/2y = 24# and #3x - 2/3y = 41/3#?

Question

Avery Alexander · Accepted Answer

#{(x = 5), (y = 2) :}#

You can solve this system of equations by multiplication.

Start by rewriting your two equations so that you can work without denominators.

#{(10x - y = 48), (9x - 2y = 41):}#

Notice that you can multiply the first equation by #-2# so that you get

#10x - y = 48 | * (-2)#

#-20x + 2y = - 96#

You can now add the two equations to cancel the #y#-terms and solve for #x#. Add the left side of the equations and the right side of the equations separately to get

#-20x + color(red)(cancel(color(black)(2y))) + 9x - color(red)(cancel(color(black)(2y))) = -96 + 41#

#-11x = -55 implies x= ((-55))/((-11)) = color(green)(5)#

Use the value of #x# in either one of the two equations to find the value of #y#

#10 * 5 - y = 48#

#y = 50 - 48 = color(green)(2)#

The two solutions to this system of equations are

#{(x = 5), (y = 2) :}#

Avery Alexander · Answer

undefined To solve the system of equations:

1. Multiply both sides of the first equation by 2 to eliminate the fraction:
   $10x - y = 48$

2. Multiply both sides of the second equation by 3 to eliminate the fraction:
   $9x - 2y = 41$

3. Now, we have the system:
   $10x - y = 48$
   $9x - 2y = 41$

4. Use the method of elimination or substitution to solve for $x$ and $y$.

Let's use the method of elimination. Multiply both sides of the first equation by 2, and both sides of the second equation by 1:
   $20x - 2y = 96$
   $9x - 2y = 41$

5. Now, subtract the second equation from the first equation to eliminate $y$:
   $20x - 9x - 2y + 2y = 96 - 41$
   $11x = 55$

6. Solve for $x$:
   $x = \frac{55}{11} = 5$

7. Substitute $x = 5$ into one of the original equations to solve for $y$. Let's use the first equation:
   $10(5) - y = 48$
   $50 - y = 48$
   $y = 50 - 48$
   $y = 2$

8. So, the solution to the system of equations is $x = 5$ and $y = 2$.