Determine the value of k 3k+1 x+3y-2 0
WebFeb 24, 2024 Β· a 11 x 1 + a 12 x 2 + β¦ + a 1n x n = 0. a 21 x 1 + a 22 x 2 + β¦ + a 2n x n = 0 β¦ a m1 x 1 + a m2 x 2 + β¦ + a mn x n = 0. The above equations containing the n unknowns x 1, x 2, β¦, x n. To determine whether the above system of equations is consistent or not, we need to find the rank of the following matrix. WebFind the Value of K for Which the System of Linear Equations Has an Infinite Number of Solutions: Kx + 3y = (2k + 1), 2(K + 1)X + 9y = (7k + 1). 0 CBSE English Medium Class 10
Determine the value of k 3k+1 x+3y-2 0
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WebLet the equal (repeated) root = r Then the equation is (X - r)^2 = 0 Expanding, we get X^2 - 2rX + r^2 = 0 Equating coefficients of like powers of X, we get r^2 = k^2 and 2r = 3(k + 1) r = Β± k ... If the roots \alpha,\beta, of the equationx^2-(3k+2)x+ (7k+1) = 0, are such that 2\alpha - \beta = 1, what is the value of k? WebSolve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more.
WebDetermine the value of k so that the following pairs of equations are inconsistent (3k + 1) x + 3y -2 = 0 (k2+ 1) x + (k - 2) y - 5 =0 Solution Suggest Corrections 0 Similar questions β¦ WebIf both the roots of quadratic equation x2 β 2kx +k2 +k β 5 = 0 are less than 5 ,then what is the interval of values of k? The solution of a quadratic equation is given by the following formula x = 2aβbΒ± b2β4ac Here a = 1, b = β2k, c = k2 + k β5 Plugging these into the formula yields... x = 22kΒ± 4k2β4(k2+kβ5) ...
WebApr 26, 2024 Β· Find an answer to your question Determine the value of k so that the following linear equations have a unique solution:(3k+1)x+3y-2=0(kΒ²+1)x+(k-2)y-5=0. Ryla Ryla 27.04.2024 Math Secondary School answered β’ expert verified WebMay 23, 2024 Β· Check the answer of values of x and y before finding value of k. x + 3y +2 = 0. 1 + 3( β 1) +2 = 0. 1 β 3 + 2 = 0. β2 + 2 = 0 ------> so values of x and y are correct. β¦
WebGiven : (3k+1)x + 3y - 2 = 0 ; (k^2+1)x+(kβ2)yβ5=0 For a system which has no solution, a1/a2 = b1/b2 β c1/c2 β΄3k+1/(k^2+1) = 3/(kβ2) β 2/5 β΄3k+1/(k^2+1) = 3/(kβ2) β¦
WebGiven equation is y=(3k+1)x 2+2(k+1)x+1=0. Also, it is given that the equation has equal roots. Then D=0βb 2β4ac=0. a=3k+1 b=2(k+1) c=1. b 2β4ac=[2(k+1)] 2β4(3k+1)=0. β¦ iowa girls basketball 2023WebFor what value of k does the pair of equations 5x+2y=2k and 2(k+1)x+ky=3k+4 have an infinite number of solutions? op eds to readWebJul 8, 2024 Β· rd sharma class 10 ch-3 pair of linear equations eg 15 page 3.71 (cbse 2001-9) oped shoe liftWebOct 10, 2024 Β· (3k+1)x+3yβ2=0 (k . 2 +1)x+(kβ2)yβ5=0. This is of the form a . 1. x+b . 1. y+c . 1 =0. a . 2. x+b . 2. y+c . 2 =0, where, a . 1 =3k+1,b . 1 =3,c . 1 =β2 . and a . 1 =k . 2 β¦ op ed support ukraineWebThe given system of equations is. (3k + 1) x + 3y β 2 = 0. (k 2 + 1) x + (k β 2) y β 5 = 0. Here, a 1 = 3k + 1, b 1 = 3, c 1 = β2. a 2 = k 2 + 1, b 2 = k β 2, c 2 = β5. The given β¦ oped supro shoulder softWebQ: If (2 ,0) is a solution of the linear equation 2x + 3y = k, then the value of k is Π 2 A: Given: (2,0) is a solution of the equation 2x+3y=k. So, x=2 and y=0. op ed telegraphWebThe discriminant of ax2 +bx+c is b2 β4ac. In your case a = (k β1), b = β2(k β1) and c = β(3k +1). Thus you need to check if the following ... 3x3+2x2-141x+280=0 Three solutions were found : x = 7/3 = 2.333 x = 5 x = -8 Step by step solution : Step 1 :Equation at the end of step 1 : ( ( (3 β’ (x3)) + 2x2) - 141x) + 280 = 0 Step 2 ... op ed the american dream is dead