
Solve the simultaneous equations: 5x + y =21 and x - 3y = 9.
Rearrange x - 3y = 9 to make x the subject by adding 3y to both sides of the equation, giving x = 9 + 3y. Substitute the value of "x" in the second equation for "9 + 3y", giving: 5(9 + 3y) + y = 21. …
Solve these simultaneous equations 5x + y = 21, x- 3y = 9
rearrange the first equation so it reads y= 5x+y= 21 y=21-5xSubstitute this y value into the second equation x-3y=9 y=21-5x x-3(21-5x)=9Expand out the bracket and rearrange to find out what x …
Solve the simultaneous equations: 5x + y = 21 and x - 3y = 9
Then add y to each side to give 21 = 45 + 16y. Then take away 45 from each side to give -24 = 16y. Now divide by 16 to give -24/16 = y. This simplifies to y = -3/2 which is the first part of the …
Solve the simultaneous equations 5x+y=21 and x-2y=9 - MyTutor
Then, to get x, divide through by 4 for x=3. That’s one half of the answer, now to get y, sub x=4 into one of the equations and rearrange for y. Subbing it into the top one: 5(4) + y = 21 so …
Solve the simultaneous equations: 5x + y = 21, x - 3y = 9
Solve the simultaneous equation, 3x + y = 8 and x + 3y = 12, to find a value for x and y. Answered by Grace R. Solve for x and y, with x and y satisfying the equations 3x+2y=36and 5x+4y=64
factorise x^2+10x+21 - MyTutor
factorise x^2+10x+21 firstly the largest power is x^2 so when factorising we know that both brackets must involve an x plus or minus a number. next we look at the 21 and we need two …
Solve the simultaneous equations: 2x-3y = 24 and 6x+2y = -5
Now we can substitute the value for y into the first equation and find x (substituting into the 2nd equation would also work fine): 2x - 3(-7) = 24 2x + 21 = 24
x is an integer such that 1≤x≤9, Prove that 0.(0x)recurring=x/99
r=0.0. x.. r=0.0x0x0x0x.... 100r=x.0x0x0x (1) 10,000r=x0x.0x0x0x (2) (2) - (1): 9,900r=x00. r=x00/9,990 r=x/99
Factorise 5 – 10m | MyTutor
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