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Algebra Linear Boldrini Exercicios Resolvidos Exercise 1. Find the value of y. 2x + 3y - x = 0 2x + 4y = 3 2x + 2y -5 = 0?Solution: x=1, y=3.5. Since the signs are reversed in this equation, it can be reduced to X+Y=0?O ?X-Y=-1?O ?X/Y=3/4?O ?X*Y=(1)*(3) O ?X^2+Y^2=(1)*(3). Exercise 2. Solve the following system: x + y = 3 2x - 5y = -11 ? ?Solution: By direct substitution we can solve ? and ? with: x=1+5 y=-1/2, x=3-5,-y=-1/3). As for ?? or ??, we have no information about the value of x and y. So we must try using elimination. Let x=1, y=3.5-? : Solution 1: 2x - 5y = -11, use elimination. Begin with the equation x + y = 3(1 + 5) : 2x = 3(1) + 5( 1) 2x=20+15=35 Solve 35 with 1 + ? : 35-1=-34 Solution 2: Since the equation is linear, use the gradient rule to find ? and ? . Let x equal 1, let y equal 3.5,-and -11/2 : ?=-34/3.5, ?=-34/3.5*sqrt(13/6)=0. 41 ?=-34/3.5*sqrt(13/6)*sqrt(4)=0.75 Solution 3:Solution: x=1, y=3.5-? : 2x = 3(1) + 5( 1)-/-11=-20+15=35.Solve 35 with 1 + ? : 35-1=-34 = 39/15. Solve -39/15 = 34which gives 1+ ? . Exercise 3. Solve the following system:Solution:: By direct substitution we can solve ? , and using elimination we can solve ?? , and using the gradient rule we can solve ?? . Exercise 4. Solve the following system:Solution: By direct substitution we can solve ? and ? .? Since we have no information about ?? , we must use the gradient rule to solve it. Solution 1: Let x=1, y=-3: 2x + 3y = 0 : 3y/2 + 4y = 0 : 3y=0+4y , y = 1. So we can conclude that there is no possible set of values for x and y so the system cannot be solved, hence it is inconsistent. cfa1e77820