How to find the solution to an equation - GRE Quantitative Reasoning
Card 0 of 768
If 6_x_ = 42 and xk = 2, what is the value of k?
If 6_x_ = 42 and xk = 2, what is the value of k?
Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.
Solve the first equation for x by dividing both sides of the equation by 6; the result is 7. Solve the second equation for k by dividing both sides of the equation by x, which we now know is 7. The result is 2/7.
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If 4_x_ + 5 = 13_x_ + 4 – x – 9, then x = ?
If 4_x_ + 5 = 13_x_ + 4 – x – 9, then x = ?
Start by combining like terms.
4_x_ + 5 = 13_x_ + 4 – x – 9
4_x_ + 5 = 12_x_ – 5
–8_x_ = –10
x = 5/4
Start by combining like terms.
4_x_ + 5 = 13_x_ + 4 – x – 9
4_x_ + 5 = 12_x_ – 5
–8_x_ = –10
x = 5/4
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If 3 – 3_x_ < 20, which of the following could not be a value of x?
If 3 – 3_x_ < 20, which of the following could not be a value of x?
First we solve for x.
Subtracting 3 from both sides gives us –3_x_ < 17.
Dividing by –3 gives us x > –17/3.
–6 is less than –17/3.
First we solve for x.
Subtracting 3 from both sides gives us –3_x_ < 17.
Dividing by –3 gives us x > –17/3.
–6 is less than –17/3.
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Quantity A
Quantity B
Quantity A
Quantity B
In order to solve for y, place x in terms of y in the first equation and then substitute that for x in the second equation.
The first equation would yield: 
.
Substituting into the second equation, we get: 
.
Simplify: 
In order to solve for y, place x in terms of y in the first equation and then substitute that for x in the second equation.
The first equation would yield:
Substituting into the second equation, we get:
Simplify:
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If 
, then, in terms of 
, 
If
You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.
You can solve this problem by plugging in random values or by simply solving for k. To solve for k, put the s values on one side and the k values on the other side of the equation. First, subtract 4s from both sides. This gives 4s – 6k = –2k. Next, add 6k to both sides. This leaves you with 4s = 4k, which simplifies to s=k. The answer is therefore s.
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What is the value of (5 + x)(10 – y) when x = 3 and y = –3?
What is the value of (5 + x)(10 – y) when x = 3 and y = –3?
This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.
This is a simple plug-in and PEMDAS problem. First, plug in x = 3 and y = –3 into the x and y. You should follow the orders of operation and compute what is within the parentheses first and then find the product. This gives 8 * 13 = 104. The answer is 104.
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If x = 4, and y = 3x + 5, then 2y – 1 equals
If x = 4, and y = 3x + 5, then 2y – 1 equals
Start by plugging in x = 4 to solve for y: y = 3 * 4 + 5 = 17. Then 2 * 17 – 1 = 33
Start by plugging in x = 4 to solve for y: y = 3 * 4 + 5 = 17. Then 2 * 17 – 1 = 33
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Sarah’s current age is three times Ron’s age two years ago. Sarah is currently 14 years older than Ron. What is the sum of Sarah and Ron’s current age?
Sarah’s current age is three times Ron’s age two years ago. Sarah is currently 14 years older than Ron. What is the sum of Sarah and Ron’s current age?
The best way to solve this problem is to turn the two statements into equations calling Sarah’s age S and Ron’s age R. So, S = 3(R – 2) and S = 14 + R. Now substitute the value for S in the second equation for the value of S in the first equation to get 14 + R = 3(R – 2) and solve for R. So R equals 10 so S equals 24 and the sum of 10 and 24 is 34.
The best way to solve this problem is to turn the two statements into equations calling Sarah’s age S and Ron’s age R. So, S = 3(R – 2) and S = 14 + R. Now substitute the value for S in the second equation for the value of S in the first equation to get 14 + R = 3(R – 2) and solve for R. So R equals 10 so S equals 24 and the sum of 10 and 24 is 34.
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A store sells potatoes for \$0.24 and tomatoes for \$0.76. Fred bought 12 individual vegetables. If he paid \$6.52 total, how many potatoes did Fred buy?
A store sells potatoes for \$0.24 and tomatoes for \$0.76. Fred bought 12 individual vegetables. If he paid \$6.52 total, how many potatoes did Fred buy?
Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.
Solving for P will give the answer.
Set up an equation to represent the total cost in cents: 24P + 76T = 652. In order to reduce the number of variables from 2 to 1, let the # tomatoes = 12 – # of potatoes. This makes the equation 24P + 76(12 – P) = 652.
Solving for P will give the answer.
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Kim is twice as old as Claire. Nick is 3 years older than Claire. Kim is 6 years older than Emily. Their ages combined equal 81. How old is Nick?
Kim is twice as old as Claire. Nick is 3 years older than Claire. Kim is 6 years older than Emily. Their ages combined equal 81. How old is Nick?
The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.
Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81
Solving for x gives Claire’s age, which can be used to find Nick’s age.
The goal in this problem is to have only one variable. Variable “x” can designate Claire’s age.
Then Nick is x + 3, Kim is 2x, and Emily is 2x – 6; therefore x + x + 3 + 2x + 2x – 6 = 81
Solving for x gives Claire’s age, which can be used to find Nick’s age.
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If 6h – 2g = 4g + 3h
In terms of g, h = ?
If 6h – 2g = 4g + 3h
In terms of g, h = ?
If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.
If we solve the equation for b, we add 2g to, and subtract 3h from, both sides, leaving 3h = 6g. Solving for h we find that h = 2g.
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If 2x + y = 9 and y – z = 4 then 2x + z = ?
If 2x + y = 9 and y – z = 4 then 2x + z = ?
If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).
The y’s cancel leaving us with an answer of 5.
If we solve the first equation for 2x we find that 2x = 9 – y. If we solve the second equation for z we find z = –4 + y. Adding these two manipulated equations together we see (2x) + (y) = (9 – y)+(–4 + y).
The y’s cancel leaving us with an answer of 5.
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11/(x – 7) + 4/(7 – x) = ?
11/(x – 7) + 4/(7 – x) = ?
We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.
We must find a common denominator and here they changed the first fraction by removing a negative from the numerator and denominator, leaving –11/(7 – x). We add the numerators and keep the same denominator to find the answer.
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Jack has 14 coins consisting of nickels and dimes that total \$0.90. How many nickels does Jack have?
Jack has 14 coins consisting of nickels and dimes that total \$0.90. How many nickels does Jack have?
In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.
For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).
Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.
When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.
In order to solve this question we must first set up two equations. We know the number of nickels and the number of dimes equals 14 (n + d = 14). We also know the value of nickels and dimes.
For the second equation we simply multiply the number of nickels we have by their value, added to the number of dimes we have by their value to get the total (0.05n + 0.10d = 0.90).
Solve the first equation for n giving us n = 14 – d. We can then substitute 14 – d into the second equation wherever there is an “n”. Giving us 0.05 (14 – d) + 0.10d = 0.90.
When we solve the equation we find the number of dimes is d = 4; therefore the remaining 10 coins must be nickels.
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If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
If a = 1/3b and b = 4c, then in terms of c, a – b + c = ?
To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c.
Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c.
To begin we must find how a and c relate to each other. Using the second equation we know that we can plug in 4c everywhere there is a b in the first equation, giving us a = 4/3c.
Now we can plug into the last equation. We plug in 4/3c for a, 4c for b, and leave c as it is. We must find a common denominator (4/3c – 12/3c + 3/3c) and add the numerators to find that our equation equals –5/3c.
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If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
If x3 = 8, then x2(4/(3 – x))(2/(4 – x)) – (4/x2) = ?
There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).
There is really no need to alter this equation using algebra. Simply find that x = 2 and plug in. We see that 4(4)(1) – (1)=15. Remember to use correct order of operations here (parentheses, exponents, multiplication, division, addition, subtraction).
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x2 + 5x – 24 = 0
y2 – 9y + 20 = 0
Quantity A
x
Quantity B
y
x2 + 5x – 24 = 0
y2 – 9y + 20 = 0
Quantity A
x
Quantity B
y
If x2 + 5x – 24 = 0,
(x – 3)(x + 8) = 0 or x = –8 or +3.
y2 – 9y + 20 = 0, then
(y – 5)(y – 4) = 0, or y = +4 or +5.
y is always greater than x.
If x2 + 5x – 24 = 0,
(x – 3)(x + 8) = 0 or x = –8 or +3.
y2 – 9y + 20 = 0, then
(y – 5)(y – 4) = 0, or y = +4 or +5.
y is always greater than x.
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One of the roots of the equation x2 + kx - 12 = 0 is 3, and k is a constant.
Quantity A: The value of k
Quantity B: -1
One of the roots of the equation x2 + kx - 12 = 0 is 3, and k is a constant.
Quantity A: The value of k
Quantity B: -1
We can factor the equation x2 + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.
x - 3 = 0
x = 3
We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.
This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.
Quantity A is greater.
We can factor the equation x2 + kx - 12 = 0, knowing that we will have (x - 3) as one of the parentheses since the root is equal to 3.
x - 3 = 0
x = 3
We also know that the other root will be -4, because we multiply the 4 and -3 in (x + 4)(x - 3) to get our constant, -12.
This means that kx is equal to 4x - 3x = x. Therefore k = 1, and quantity A > quantity B.
Quantity A is greater.
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y = x2 - 10
y = 15
Quantity A: y/3
Quantity B: x
y = x2 - 10
y = 15
Quantity A: y/3
Quantity B: x
We know that Quantity A = y / 3 = 15 /3 = 5.
If we plug in 15 for y, we can solve for x, for Quantity B.
y = x2 - 10
y = 15
15 = x2 - 10 (Add 10 to both sides.)
25 = x2
x = 5 or -5
Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.
We know that Quantity A = y / 3 = 15 /3 = 5.
If we plug in 15 for y, we can solve for x, for Quantity B.
y = x2 - 10
y = 15
15 = x2 - 10 (Add 10 to both sides.)
25 = x2
x = 5 or -5
Since 5 is equal to 5 but is greater than -5, we cannot determine the relationship between Quantities A and B.
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Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
Find the intersection of the following two equations:
3x + 4y = 6
15x - 4y = 3
The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:
3x + 4y = 6
15x - 4y = 3
18x = 9; x = 0.5
You can now plug x into the first equation:
3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125
Therefore, our point of intersection is (0.5, 1.125)
The point of intersection for two lines is the same as the values of x and y that mutually solve each equation. Although you could solve for one variable and replace it in the other equation, use elementary row operations to add the two equations since you have a 4y and -4y:
3x + 4y = 6
15x - 4y = 3
18x = 9; x = 0.5
You can now plug x into the first equation:
3 * 0.5 + 4y = 6; 1.5 +4y = 6; 4y = 4.5; y = 1.125
Therefore, our point of intersection is (0.5, 1.125)
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