How do you solve #4x+7y=6# and #6x+5y=20# using elimination?

Question

Ethan Adkins · Accepted Answer

#{(6x+5y=20, 4x+7y=6):}#

the first equation by #3# and the second equation by #2# can be multiplied.

#=>{12x + 21 y = 18, 12x + 10 y = 40]:}#

The second equation is subtracted from the first equation.

#=>11y=-22#

By dividing #11# by 2,

#=>y=-2#

By entering #y=-2# into the original system's first equation,

#4x-14=6# since #4x+7(-2)=6.

By incorporating #14#,

#=> 4x=20#

Taking the division by #4#,

#=> x = 5#

For this reason, #(x,y)=(5,-2)# is the solution.

I hope you found this useful.

Hazel Anglin · Answer

undefined To solve the system of equations $4x + 7y = 6$ and $6x + 5y = 20$ using elimination, you can multiply each equation by a suitable constant to make the coefficients of one of the variables the same or additive inverses. Multiplying the first equation by 5 and the second equation by 7, we get:

$20x + 35y = 30$ (equation 1 multiplied by 5)
$42x + 35y = 140$ (equation 2 multiplied by 7)

Now, subtract equation 1 from equation 2:

$(42x + 35y) - (20x + 35y) = 140 - 30$
$42x + 35y - 20x - 35y = 110$
$22x = 110$

Divide both sides by 22:

$x = \frac{110}{22}$
$x = 5$

Now, substitute the value of $x$ into one of the original equations (e.g., the first one) and solve for $y$:

$4(5) + 7y = 6$
$20 + 7y = 6$
$7y = 6 - 20$
$7y = -14$
$y = -2$

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