How to solve systems by substitution?
Let q be the number of quarters and d be the number of dimes. You know there are 8 coins total, so that means q + d = 8. Since a quarter is equivalent to $0.25, then 0.25q is the amount of money made up from the quarters by themselves; likewise, the dime is equivalent to $0.10, so 0.1d is the amount of money made up from the dimes by themselves. And so the total money is 0.25q + 0.1d = 1.25. Therefore, your system of equations is: q + d = 8 0.25q + 0.1d = 1.25 Now solve by substitution. First solve for one of the variables in one of the equations. I'll solve for d in the first equation: q + d = 8 (subtract q from both sides) d = 8 - q Now you can substitute 8 - q for d into the second equation in order to solve for q: 0.25q + 0.1d = 1.25 (substitute 8 - q for d) 0.25q + 0.1(8 - q) = 1.25 (distribute 0.1) 0.25q + 0.8 - 0.1q = 1.25 (combine like terms) 0.15q + 0.8 = 1.25 (subtract 0.8 from both sides) 0.15q = 0.45 (divide both sides by 0.15) q = 3 And now go back to solve for d: d = 8 - q (substitute 3 for q) d = 8 - 3 d = 5 ANSWER: Maribel has 5 dimes and 3 quarters.
Solve the system by substitution?
2.5x - 3y = -13 (1) 3.25x - y = -14 (2) From (2), y = 3.25x + 14 Substitute y = 3.25x + 14 into (1): 2.5x - 3(3.25x + 14) = -14 2.5x - 9.75x - 42 = -14 7.25x = -29 x = -4 Substitute x = -4 into (2): y = 3.25(-4) + 14 = 1 Therefore x = -4 and y = 1.
Math problem solving systems by substitution?
Ok first you have to take one of your equations and solve for a variable of your choice, im going to choose solving for Y in the second equation since it looks the easiest... So you end up with Y=16X-18 Then subsitute that in for the first equation so it becomes... 0.5X+0.25(16X-18)=36 Then multiply out the equation so there is no perenthese 0.5X+4X-4.5=36 Then solve for X now that there is only X's 4.5X=40.5 X=9 Now that you know what X equals subsitute it into one of your equations and solve for Y.. Y+18= 16(9) Y+18=144 Y=126
Solving systems by substitution and elimination?
Solve the system by substitution. -4.5 - 2y = -12.5 3.25x - y = -0.75 ---------------------------------------... Solve the system by elimination 5x + 4y = 12 3x - 3y = 18
Please HELP! Solving systems of linear systems by substitution method #33?
An investment of $3500 is divided between two simple interest accounts. On one account, the annual simple interest rate is 5% and on the second account, the annual simple interest rate is 7.5%. How much should be invested in each account so that the total interest from the two accounts is $215? I know the answer is the amounts invested should be $1900 at 5% and $1600 at 7.5% but how do I get the answer?? Your help is greatly appreciated, I am so stuck on these questions!
Please HELP! Solving systems of linear systems by substitution method #37?
A police officer has chosen a high-yield fund that earns 8% annual simple interest for part of a $6000 investment. The remaining portion is used to purchase a preferred stock that earns 11% annual simple interest. How much should be invested in each account so that the amount earned on the 8% account is twice the amount earned on the 11% account? I know the answer is the amounts invested should be $4400 at 8% and $1600 at 11% but how do I get the answer?? Your help is greatly appreciated, I am so stuck on these questions!
Solving systems by substitution (with work please)?
1. y=2x+8 and y=8 put the value of y = 8 in first , we have 8 = 2x + 8 => 2x =0 => x = 0 and y = 8 2. x+y=-3 and x-y=-1 just add both the equations ( x+y )+ (x-y) = (-3) + (-1) => 2x = -4 => x =-2 because x+y=-3 => y = - 3 - x => y = -3 -(-2) => y = -1 so x =-2 and y = -1 3.4x+2y=-2 and y=6x-5 multiply second by 2 and subtract from first (4x + 2y) - 2y = - 2 -12x + 10 => 16x = 8 => x =1/2 also y = 6x -5 y = 6(1/2) -5 = 3- 5 = -2 => y = -2 4. 1/2x=y-1 and 1/3x=y multiply first by 2 and second by three x = 2y - 2 , x =3y so 3y = 2y - 2 => y = -2 so x = 3y => x = 3(-2) = -6 5 .x=-1/4y+5 and 3x+2y=0 multiply first by 8 8x + 2y = 40 subtract both equations , (8x + 2y ) - (3x+2y) =40 5x = 40 => x =8 also y = -3/2x => y = -3x8/2 = -12 6.3x=-y+10 and 2x+3y=-12 multiply first by 3 . 9x + 3y = 30 and 2x +3y = -12 subtract both , ( 9x + 3y ) - ( 2x +3y ) = 30 - (-12) 7x = 42 => x = 6 also 2x +3y = -12 so 12 + 3y = -12 => 3y = -24 => y = -8
How do you solve systems of equations by substitution?
Consider the following system of linear equations:-2x-y=8…….(i)& 3x+y=9……(ii)Then we've to solve it by substitution,Step 1:- find the value of either x or y from any two of the equation.i.e. from equation (i) ,2x-y=8=> y= 2x-8…..(¡¡¡)Step2:- Put the value of y in equation (ii) to get the value of x.i.e. 3x+y=9=> 3x+(2x-8)=9=> 5x-8=9=> 5x=9+8=17=> x= 17/5Step3:- Put the value of x in equation (iii) to get the value of y.i.e. y=2x-8=> y=(2×17/5)-8=> y=(34/5)-8=(34–40)/5=> y=-6/5Hence, the corresponding values of x & y are17/5 & -6/5 respectively……. Ans.
How would you solve 3y = 6x - 5 and y = 2x - 5 using the substitution method?
We can solve the given system of two linear equations in two variables: 3y = 6x - 5 and y = 2x - 5 by using the Substitution Method as follows:Because the second equation already states that y = 2x - 5, we can substitute 2x - 5 in for y in the first equation as follows:3y = 6x - 53(2x - 5) = 6x - 53(2x) - 3(5) = 6x - 56x - 15 = 6x - 5Now, add 15 to both sides of the equation as follows:6x - 15 + 15 = 6x - 5 + 156x + 0 = 6x + 106x = 6x + 10Now, subtract 6x from both sides:6x - 6x = 6x - 6x + 100x = 0x + 100 = 10 which is a false statement!Therefore, what we have is a system of two linear equations in two unknowns which has NO solution. Geometrically, to analyze further to see why there is no solution, the first equation can also be written in slope-intercept form (y = mx + b) as: y = 2x - (5/3), and we now see from the two equations that the graphs (lines) of the two equations have the same slope: m = 2; therefore, we see that the graphs of the two given equations are actually parallel lines which, as we know, are lines in a plane which will never intersect, i.e., they will never have a point in common, the coordinates (x, y) of which, if it existed, would be the solution to the given system. Consequently, the given system of equations has NO solution; So, what we have is what's called an inconsistent system.