#1**+1 **

Solve the following system:

{2 x + 4 y = -32

y - 3 x = 6

In the second equation, look to solve for y:

{2 x + 4 y = -32

y - 3 x = 6

Add 3 x to both sides:

{2 x + 4 y = -32

y = 3 x + 6

Substitute y = 3 x + 6 into the first equation:

{2 x + 4 (3 x + 6) = -32

y = 3 x + 6

2 x + 4 (3 x + 6) = (12 x + 24) + 2 x = 14 x + 24:

{14 x + 24 = -32

y = 3 x + 6

In the first equation, look to solve for x:

{14 x + 24 = -32

y = 3 x + 6

Subtract 24 from both sides:

{14 x = -56

y = 3 x + 6

Divide both sides by 14:

{x = -4

y = 3 x + 6

Substitute x = -4 into the second equation:

**Answer: | x = -4 and y = -6**

Guest Jul 12, 2017

#2**+2 **

To solve for *x *and *y, *first solve for a variable and plug it into the other equation. Since you specified solving by substitution, I'll use that method. Here are your 2 equations:

1. \(2x+4y=-32\)

2. \(-3x+y=6\)

I'll solve for *y *in the second equation because it has a coefficient of 1.

\(-3x+y=6\) | To isolate y, add 3x to both sides. |

\(y=6+3x\) | |

Now that I have solved for *y *in one equation, substitute y in the other.

\(2x+4y=-32\) | In the previous calculation, we deduced that y=6+3x, so replace y. |

\(2x+4(6+3x)=-32\) | Distribute the 4 into the parentheses to ease the simplification process. |

\(2x+24+12x=-32\) | Combine the like terms, specifically 2x and 12x. |

\(14x+24=-32\) | Subtract 24 on both sides. |

\(14x=-56\) | Divide by 14 on both sides to isolate x. |

\(x=-4\) | |

Plug *x *into an equation and solve for *y*. I'll plug it into equation 2

\(-3x+y=6\) | Substitute the calculated value for x, -4. |

\(-3(-4)+y=6\) | Simplify -3*-4. |

\(12+y=6\) | Subtract 12 on both sides |

\(y=-6\) | |

Therefore, the solution set is (-4,-6)

TheXSquaredFactor Jul 12, 2017

#3**0 **

**x+y=97; x-y=39 / using elimination method/ SHOW WORK**

Guest Jul 12, 2017

edited by
Guest
Jul 12, 2017

#4**0 **

Solve the following system:

{x + y = 97 | (equation 1)

x - y = 39 | (equation 2)

Subtract equation 1 from equation 2:

{x + y = 97 | (equation 1)

0 x - 2 y = -58 | (equation 2)

Divide equation 2 by -2:

{x + y = 97 | (equation 1)

0 x+y = 29 | (equation 2)

Subtract equation 2 from equation 1:

{x+0 y = 68 | (equation 1)

0 x+y = 29 | (equation 2)

Collect results:

**Answer: | x = 68 and y = 29**

**Please post your question in the "Questions", NOT in the "Answers" section.**

Guest Jul 12, 2017