You can certainly start out that way, but when you solve that equation for x, it's a little more complex than if you started with the other equation. I usually start with the most simple equation first. Let's take the first equation and solve for y... that will leave us without any fractions to deal with.
2x + y = -14
y = -14 - 2x
Now we have a y value and we can substitute it back into the other equation. So we will replace y in the second equation with (-14 - 2x).
That will leave us with just the x as a variable and we can solve for x.
7x - 4y = -19
7x - 4(-14 - 2x) = -19 multiply the -4 through the parentheses
7x + 56 + 8x = -19 combine like terms
15x + 56 = -19 subtract 56 from each side
15x = -75 divide each side by 15
x = -5
Now we have a value for x that we can substitute back into either of the original equations and then solve for y. I usually go with the easier equation, but it doesn't matter. Let's use the first one...
2x + y = -14
2(-5) + y = -14 multiply the 2 through the parentheses
-10 + y = -14 add 10 to each side
y = -4
So your ordered pair is
(-5, -4)
That is where the 2 lines are equal to one another, so that's the point where they they intersect.
Note*** You can start these problems with either equation and solving for either x or y... it doesn't matter. After you substitute the values into the other equation it will work out the same.
Answer:
4)
Step-by-step explanation:hope this helps
Answer:
Answer for 4th question
b = 7
Answer for 5th question
area= 99sq.in
Step-by-step explanation:
4. area of parallelogram= base × height
28= b × 4
b= 28/4
b = 7
5. area of parallelogram= base × height
area= 9 × 11
area= 9 sq.in
Answer:
b
Step-by-step explanation:
Answer:
It is impossible for a number to be both less than 1 and greater than or equal to 3, so the compound inequality has NO SOLUTION.
Step-by-step explanation:
The word "and" tells us that we are looking for all values of x that satisfy BOTH inequalities simultaneously.
3x + 5 < 8 and 2x + 5 ≥ 11
3x < 3 and 2x ≥ 6
x < 1 and x ≥ 3
It is impossible for a number to be both less than 1 and greater than or equal to 3, so the compound inequality has NO SOLUTION.