Question

PleaseeeeeHelp!!!!!!!!!!!!

Answer 1

Answer:

x = -2, y = -5

Step-by-step explanation:

Isolate x for 3x - 2y = 4

3x - 2y = 4

Add 2y to both sides

3x -2y + 2y = 4 + 2y

Simplify

3x = 4 + 2y

Divide both sides by 3

$\frac{3x}{3} =\frac{4}{3} +\frac{2y}{3}$

Simplify

$x=\frac{4+2y}{3}$

Now we substitute  $x=\frac{4+2y}{3}$  for x in 4x - 3y = 7

$4*\frac{4+2y}{3}-3y=7$

Isolate y for $4*\frac{4+2y}{3}-3y=7$

$4*\frac{4+2y}{3}-3y=7$

Expand $4*\frac{4+2y}{3}-3y$

$4*\frac{4+2y}{3} =\frac{16+8y}{3}$

$4*\frac{4+2y}{3}$

Multiply fractions: $a*\frac{b}{c} =\frac{a*b}{c}$

$=\frac{(4+2y)*4}{3}$

Expand $(4+2y)*4$

$=4(4+2y)$

Apply the distributive law: $a(b+c)=ab+ac$

$a=4, b=4,c=2y$

$=4*4+4*2y$

Simplify $4*4+4*2y$

Multiply the numbers: $4*4 = 16$

$=16+4*2y$

Multiply the numbers: $4*2=8$

$=16+8y$

$=\frac{16+8y}{3}$

$=\frac{8y+16}{3}-3y$

Convert element to fraction: $3y=\frac{3y3}{3}$

$=\frac{16+8y}{3} -\frac{3y*3}{3}$

Since the denominators are equal, combine the fractions: $\frac{a}{c}$±$\frac{b}{c}= \frac{a±b}{c}$

$=\frac{16+8y-3y*3}{3}$

$16+8y-3y*3$

Multiply the numbers: $3*3=9$

$=16+8y-9y$

Add similar elements: $8y-9y=-y$

$=16-y$

$=\frac{16-y}{3}$

Apply the fraction rule: $\frac{a}{c}$±$\frac{b}{c}= \frac{a±b}{c}$

$\frac{16-y}{3} =\frac{16}{3} -\frac{y}{3}$

$=\frac{16}{3}-\frac{y}{3}$

$\frac{16}{3} -\frac{y}{3} =7$

Multiply both sides by 3

$\frac{16}{3}*3-\frac{y}{3} *3=7*3$

Simplify

$16-y=21$

Subtract 16 from both sides

$16-y-16=21-16$

Simplify

$-y=5$

Divide both sides by -1

$\frac{-y}{-1} =\frac{5}{-1}$

Simplify

$y=-5$

For $x=\frac{4+2y}{3}$ substitute $y=-5$

$x=\frac{4+2y(-5)}{3}$

Remove parentheses: $(-a)=-a$

$=\frac{4+2y*5}{3}$

Multiply the numbers: $2*5=10$

$=4-10$

Subtract the numbers: $4-10=-6$

$=-6$

$=\frac{-6}{3}$

Apply the fraction rule: $\frac{-a}{b} =-\frac{a}{b}$

$=-\frac{6}{3}$

Divide the numbers: $\frac{6}{3}=2$

$=-2$

The solutions to the system of equations are

$y=-5,x=-2$

Checking answers

Plug in $x=-2$ and $y=-5$ into $3x-2y=4$ and $4x-3y=7$

$3(-2)-2(-5)=4$

Remove parentheses: $(-a)=-a,-(-a)=a$

$=3*2+2=5$

Multiply the numbers: $3*2=6$

$=-6+2*5$

Multiply the numbers: $2*5=10$

$=-6+10$

Add/subtract the numbers: $-6+10=4$

$=4$

First equation proven true

Substitute the values of x and y into the second equation

$4(-2)-3(-5)=7$

Follow the PEMDAS order of operations

Multiply and divide left to right $4(-2)$

$4(-2)$

Apply rule $a*(-b)=-a*b$

$4(-2)=-4*2=-8$

$=-8$

$=-8-3(-5)$

Multiply and divide left to right $3(-5)$

$3(-5)$

Apply rule $a*(-b)=-a*b$

$3*(-5)=-3*5=-15$

$=-15$

$=-8-(-15)$

Add and subtract left to right $-8-(-15)$

$-8-(-15)$

Apply rule $-(-a)=+a$

$-(-15)=+15$

$=-8+15$

$-8+15=7$

$=7$

Second equation proven true.

Both equations are true with $x=-2$ and  $y=-5$