A) AA B) SAS C) SSS D) NOT
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Solve the following using Substitution method
2x – 5y = -13
3x + 4y = 15
- To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.
- Choose one of the equations and solve it for x by isolating x on the left-hand side of the equal sign. I'm choosing the 1st equation for now.
- Add 5y to both sides of the equation.
- Divide both sides by 2.
- Multiply
times 5y - 13.
- Substitute
for x in the other equation, 3x + 4y = 15.
- Multiply 3 times
- Add
to 4y.
- Add
to both sides of the equation.
- Divide both sides of the equation by 23/2, which is the same as multiplying both sides by the reciprocal of the fraction.
- Substitute 3 for y in
. Because the resulting equation contains only one variable, you can solve for x directly.
- Multiply 5/2 times 3.
- Add
to
by finding a common denominator and adding the numerators. Then reduce the fraction to its lowest terms if possible.
- The system is now solved. The value of x & y will be 1 & 3 respectively.
Well, how long will it take you to drive 135 miles at 55mph?
at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.
so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.
so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.
at 55 mph, you're doing 55 miles every hour, so we can simply get the quotient of 135/55 and that's how many hours it'll take you to drive 135 miles at that speed.
so, it takes you that long, however, from 3:15 to 5:00pm there are only 45mins + 60mins or 1hr and 45 minutes, namely 1¾ hr.
so 2hrs and 27 minutes is much later than 1¾ hr, so, no dice, you can't arrive at 5pm, actually you'll arrive around 5:42pm.