How to find the solution to an equation - ACT Math
Card 0 of 1134
Give the lines y = 0.5x+3 and y=3x-2. What is the y value of the point of intersection?
Give the lines y = 0.5x+3 and y=3x-2. What is the y value of the point of intersection?
In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.
In order to solve for the x value you set both equations equal to each other (0.5x+3=3x-2). This gives you the x value for the point of intersection at x=2. Plugging x=2 into either equation gives you y=4.
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If a%b = (2b + 3a)/(6ab), what would have a greater value, 2%3 or 3%2?
If a%b = (2b + 3a)/(6ab), what would have a greater value, 2%3 or 3%2?
First find 2%3 = (2 * 3 + 3 * 2)/(6 * 2 * 3) = 12/36 = 1/3, then 3%2 = (2 * 2 + 3 * 3)/(6 * 3 * 2) = 13/36 which is greater.
First find 2%3 = (2 * 3 + 3 * 2)/(6 * 2 * 3) = 12/36 = 1/3, then 3%2 = (2 * 2 + 3 * 3)/(6 * 3 * 2) = 13/36 which is greater.
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If 12x + 3 = 2(5x + 5) + 1, what is the value of x?
If 12x + 3 = 2(5x + 5) + 1, what is the value of x?
Starting with 12x + 3 = 2(5x + 5) + 1, we start by solving the parenthesis, giving us 12x + 3 = 10x + 11. We then subtract 10x from the right side and subtract three from the left, giving us 2x = 8; divide by 2 → x = 4.
Starting with 12x + 3 = 2(5x + 5) + 1, we start by solving the parenthesis, giving us 12x + 3 = 10x + 11. We then subtract 10x from the right side and subtract three from the left, giving us 2x = 8; divide by 2 → x = 4.
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If you multiply two integers together and then add 5, the result is 69. Which of the following could not be the sum of the two integers?
If you multiply two integers together and then add 5, the result is 69. Which of the following could not be the sum of the two integers?
The equation is xy + 5 = 69, making xy equal to 64. If we factor 64, we see that 1x64, 2x32, 4x16 and 8x8 all equal 16 when the two numbers are added together, so 24 is the only possibility that does not work.
The equation is xy + 5 = 69, making xy equal to 64. If we factor 64, we see that 1x64, 2x32, 4x16 and 8x8 all equal 16 when the two numbers are added together, so 24 is the only possibility that does not work.
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If 5 + x is 5 more than 5,what is the value of 2_x_?
If 5 + x is 5 more than 5,what is the value of 2_x_?
5 more than 5 = 10
5 + x = 10
Subtract 5 from each side of the equation: x = 5 → 2_x_ = 10
5 more than 5 = 10
5 + x = 10
Subtract 5 from each side of the equation: x = 5 → 2_x_ = 10
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Three consecutive positive numbers have the sum of 15. What is the product of these numbers?
Three consecutive positive numbers have the sum of 15. What is the product of these numbers?
Define the variables as x = the first number, x + 1, the second number, and x + 2 the thrid number.
The sum becomes x + x + 1 + x + 2 = 15 so 3x + 3 = 15. Subtract 3 from both sides of the equation to get 3x = 12 → 3x/3 = 12/3 → x = 4
The three numbers are 4, 5, and 6 and their product is 120.
Define the variables as x = the first number, x + 1, the second number, and x + 2 the thrid number.
The sum becomes x + x + 1 + x + 2 = 15 so 3x + 3 = 15. Subtract 3 from both sides of the equation to get 3x = 12 → 3x/3 = 12/3 → x = 4
The three numbers are 4, 5, and 6 and their product is 120.
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A company makes toy boats. Their monthly fixed costs are \$1500. The variable costs are \$50 per boat. They sell boats for \$75 a piece. How many boats must be sold each month to break even?
A company makes toy boats. Their monthly fixed costs are \$1500. The variable costs are \$50 per boat. They sell boats for \$75 a piece. How many boats must be sold each month to break even?
The break-even point is where the costs equal the revenues
Fixed Costs + Variable Costs = Revenues
1500 + 50_x_ = 75_x_
Solving for x results in x = 60 boats sold each month to break even.
The break-even point is where the costs equal the revenues
Fixed Costs + Variable Costs = Revenues
1500 + 50_x_ = 75_x_
Solving for x results in x = 60 boats sold each month to break even.
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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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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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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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If A + B = C and B – C = 3, what does A equal?
If A + B = C and B – C = 3, what does A equal?
This question can be solved using substitution. There are several ways to subsitute, but one possible method is to substitue (A+B) for C in the equation B-C=3: B-(A+B)=3. This simplifies to (B-A-B=3) (don't forget to distribute the negative!). Since (B-B)=0, -A must equal 3 (-A=3). If we divide both sides by -1, A=-3. If you got an answer of positive 3, you may have forgotten to divide both sides by -1.
This question can be solved using substitution. There are several ways to subsitute, but one possible method is to substitue (A+B) for C in the equation B-C=3: B-(A+B)=3. This simplifies to (B-A-B=3) (don't forget to distribute the negative!). Since (B-B)=0, -A must equal 3 (-A=3). If we divide both sides by -1, A=-3. If you got an answer of positive 3, you may have forgotten to divide both sides by -1.
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If y=0.35(300 – 2y), what is the value of y to the nearest tenth?
If y=0.35(300 – 2y), what is the value of y to the nearest tenth?
Distributing the 0.35 to both the 300 and **–**2y leaves y=105 – 0.7y
Adding 0.7y to both sides and dividing by 1.7 gives 61.8.
Distributing the 0.35 to both the 300 and **–**2y leaves y=105 – 0.7y
Adding 0.7y to both sides and dividing by 1.7 gives 61.8.
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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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What is the value of c when 3c + 4 = 2c – 7?
What is the value of c when 3c + 4 = 2c – 7?
First put the variables on one side by subtracting 2c on both sides to get c + 4 = –7. Then subtract the 4 on both sides to arrive at your answer, c = **–**11.
First put the variables on one side by subtracting 2c on both sides to get c + 4 = –7. Then subtract the 4 on both sides to arrive at your answer, c = **–**11.
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The population of a bird species is modeled by the following equation:
,
where 
represents the number of years from the present. How many years will it take the population to reach 130 birds (rounded to the nearest tenth)?
The population of a bird species is modeled by the following equation:
where
Plugging in 130 for P, the equation becomes 130 = (11/8)x + 102. Solving for x, we obtain x = 20.4 years.
Plugging in 130 for P, the equation becomes 130 = (11/8)x + 102. Solving for x, we obtain x = 20.4 years.
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John has \$50 for soda and he must buy both diet and regular sodas. His total order must have at exactly two times as many cans of diet soda as cans of regular soda. What is the greatest number of cans of diet soda John can buy if regular soda is \$0.50 per can and diet soda is \$0.75 per can?
John has \$50 for soda and he must buy both diet and regular sodas. His total order must have at exactly two times as many cans of diet soda as cans of regular soda. What is the greatest number of cans of diet soda John can buy if regular soda is \$0.50 per can and diet soda is \$0.75 per can?
From our data, we can come up with the following two equations:
0.50R + 0.75D = 50
2R = D
Replace the D value in the second equation into the first one:
0.5R + 0.75 * 2R = 50
0.5R + 1.5R = 50; 2R = 50; R = 25
However, note that the question asks for the number of diet cans, so this will have to be doubled to 50.
From our data, we can come up with the following two equations:
0.50R + 0.75D = 50
2R = D
Replace the D value in the second equation into the first one:
0.5R + 0.75 * 2R = 50
0.5R + 1.5R = 50; 2R = 50; R = 25
However, note that the question asks for the number of diet cans, so this will have to be doubled to 50.
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